Chemistry Class 12 NCERT Solutions Chapter Solutions – Important Questions
Q 2.1) If 22 g of benzene is dissolved in 122 g of carbon tetrachloride, determine the mass percentage of carbon tetrachloride (CCl4) and benzene (C6H6).
Answer 2.1:
Mass percentage of Benzene (C6H6) =
\(\begin{array}{l}\frac{Mass\; of\; C_{6}H_{6}}{Total \;mass \;of \;the \;solution} v\times 100\end{array} \)
=
\(\begin{array}{l}\frac{Mass\; of\; C_{6}H_{6}}{Mass\;of\;C_{6}H_{6} + Mass\;of\;CCl_{4}} \times 100\end{array} \)
=
\(\begin{array}{l}\frac{22}{22 + 122}\times 100\end{array} \)
= 15.28%
Mass percentage of Carbon Tetrachloride (CCl4) =
\(\begin{array}{l}\frac{Mass\; of\; CCl_{4}}{Total \;mass \;of \;the \;solution} \times 100\end{array} \)
=
\(\begin{array}{l}\frac{Mass\; of\; CCl_{4}}{Mass\;of\;C_{6}H_{6} + Mass\;of\;CCl_{4}} \times 100\end{array} \)
=
\(\begin{array}{l}\frac{122}{22 + 122}\times 100\end{array} \)
= 84.72%
Q 2.2) If benzene in a solution contains 30% by mass in carbon tetrachloride, calculate the mole fraction of benzene.
Answer 2.2:
Assume the mass of benzene is 30 g in the total mass of the solution of 100 g.
Mass of CCl4 = (100 − 30) g
= 70 g
Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g
\(\begin{array}{l}mol^{-1}\end{array} \)
= 78 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Therefore, the number of moles of C6H6 =
\(\begin{array}{l}\frac{30}{78}\end{array} \)
mol
= 0.3846 mol
Molar mass of CCl4 = 1 x 12 + 4 x 355 = 154 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Therefore, the number of moles of CCl4 =
\(\begin{array}{l}\frac{70}{154}\end{array} \)
mol
= 0.4545 mol
Thus, the mole fraction of C6H6 is given as
\(\begin{array}{l}\frac{Number\;of\;moles\;of\;C_{6}H_{6}}{Number\;of\;moles\;of\;C_{6}H_{6} + Number\;of\;moles\;of\;CCl_{4}}\end{array} \)
=
\(\begin{array}{l}\frac{0.3846}{0.3846 + 0.4545}\end{array} \)
= 0.458
Q 2.3) Determine the molarity of each of the solutions given below:
(a) 30 g of Co(NO)3. 6H2O in 4.3 L of solution.
(b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.
Answer 2.3:
We know that,
Molarity =
\(\begin{array}{l}\frac{Moles\;of\;Solute}{Volume\;of\;solution\;in\;litre}\end{array} \)
(a) Molar mass of Co(NO)3. 6H2O = 59 + 2 (14 + 3 x 16) + 6 x 18 = 291 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Therefore, the moles of Co(NO)3. 6H2O =
\(\begin{array}{l}\frac{30}{291}\end{array} \)
mol
= 0.103 mol
Therefore, molarity =
\(\begin{array}{l}\frac{0.103\; mol}{4.3\; L}\end{array} \)
= 0.023 M
(b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol
Therefore, the number of moles present in 30 mL of 0.5 M H2SO4 =
\(\begin{array}{l}\frac{0.5\times 30}{1000}\; mol\end{array} \)
= 0.015 mol
Therefore, molarity =
\(\begin{array}{l}\frac{0.015}{0.5\; L}\; mol\end{array} \)
= 0.03 M
Q 2.4) To make 2.5 kg of 0.25 molar aqueous solution, determine the mass of urea (NH2CONH2) that is required.
Answer 2.4:
Molar mass of urea (NH2CONH2) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16 = 60 g
\(\begin{array}{l}mol^{-1}\end{array} \)
0.25 molar aqueous solution of urea means
1000 g of water contains 0.25 mol = (0.25 × 60) g of urea = 15 g of urea
That is,
(1000 + 15) g of solution contains 15 g of urea
Therefore, 2.5 kg (2500 g) of solution contains =
\(\begin{array}{l}\frac{15\times 2500}{1000 + 15}\; g\end{array} \)
= 36.95 g
= 37 g of urea (approx.)
Hence, the mass of Urea required is 37 g.
Q 2.5) If 1.202 g
\(\begin{array}{l}mL^{-1}\end{array} \)
is the density of 20% aqueous KI, determine the following:
(a) Molality of KI
(b) Molarity of KI
(c) Mole fraction of KI
Answer 2.5:
(a) Molar mass of KI = 39 + 127 = 166 g
\(\begin{array}{l}mol^{-1}\end{array} \)
20% aqueous solution of KI means 20 g of KI is present in 100 g of solution.
That is,
20 g of KI is present in (100 – 20) g of water = 80 g of water.
Therefore, molality of the solution =
\(\begin{array}{l}\frac{Moles\;of\;KI}{Mass\;of\;water\;in\;kg}\end{array} \)
=
\(\begin{array}{l}\frac{\frac{20}{166}}{0.08}\; m\end{array} \)
= 1.506 m
= 1.51 m (approx.)
(b) It is given that the destiny of the solution = 1.202
\(\begin{array}{l}g\; mL^{-1}\end{array} \)
Volume of 100 g solution =
\(\begin{array}{l}\frac{Mass}{Density}\end{array} \)
=
\(\begin{array}{l}\frac{100\; g}{1.202\;g \; mL^{-1}}\end{array} \)
= 83.19 mL
=
\(\begin{array}{l}83.19\times 10^{-3}\; L\end{array} \)
Therefore, the molarity of the solution =
\(\begin{array}{l}\frac{\frac{20}{166}\; mol}{83.19\times 10^{-3}\; L}\end{array} \)
= 1.45 M
(c) Moles of KI =
\(\begin{array}{l}\frac{20}{166}\end{array} \)
= 0.12 mol
Moles of water =
\(\begin{array}{l}\frac{80}{18}\end{array} \)
= 4.44 mol
Therefore, mole =
\(\begin{array}{l}\frac{Moles\; of\; KI}{Moles\; of\; KI + Moles\; of \; water}\end{array} \)
Fraction of KI =
\(\begin{array}{l}\frac{0.12}{0.12 + 4.44}\end{array} \)
= 0.0263
Q 2.6) Calculate Henry’s law constant when the solubility of H2S (a toxic gas with a rotten egg-like smell) in water at STP is 0.195 m
Answer 2.6:
It is given that the solubility of H2S in water at STP is 0.195 m, i.e., 0.195 mol of H2S is dissolved in 1000 g of water.
Moles of water =
\(\begin{array}{l}\frac{1000\; g}{18\; g\; mol^{-1}}\end{array} \)
= 55.56 mol
Therefore, the mole fraction of H2S, x =
\(\begin{array}{l}\frac{Moles\;of\;H_{2}S}{Moles\;of\;H_{2}S + Moles\;of\;water}\end{array} \)
=
\(\begin{array}{l}\frac{0.195}{0.195 + 55.56}\end{array} \)
= 0.0035
At STP, pressure (p) = 0.987 bar
According to Henry’s law, p =
\(\begin{array}{l}K_{H}\end{array} \)
x
=>
\(\begin{array}{l}K_{H} = \frac{P}{x}\end{array} \)
=
\(\begin{array}{l}\frac{0.987}{0.0035}\; bar\end{array} \)
= 282 bar
Q 2.7) A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Answer 2.7:
The total amount of solute present in the mixture is given by,
\(\begin{array}{l}300\times \frac{25}{100} + 400\times \frac{40}{100}\end{array} \)
= 75 + 160
= 235 g
Total amount of solution = 300 + 400 = 700 g
Therefore, mass percentage of the solute in the resulting solution =
\(\begin{array}{l}\frac{235}{700}\times 100\end{array} \)
= 33.57%
And the mass percentage of the solvent in the resulting solution is
= (100 – 33.57) %
= 66.43%
Q 2.8) The vapour pressure of pure liquids A and B are 450 and 700 mm Hg, respectively, at 350 K. Find out the composition of the liquid mixture if the total vapour pressure is 600 mm Hg. Also, find the composition of the vapour phase.
Answer 2.8:
It is given that
\(\begin{array}{l}P^{\circ}_{A}\end{array} \)
= 450 mm of Hg
\(\begin{array}{l}P^{\circ}_{B}\end{array} \)
= 700 mm of Hg
\(\begin{array}{l}P_{total}\end{array} \)
= 600 mm of Hg
According to Raoult’s law,
\(\begin{array}{l}P_{A} = P^{\circ}_{A}x_{A}\end{array} \)
\(\begin{array}{l}P_{B} = P^{\circ}_{B}x_{B} = P^{\circ}_{B}(1 – x_{A})\end{array} \)
Therefore, total pressure,
\(\begin{array}{l}P_{total} = P_{A} + P_{B}\end{array} \)
=>
\(\begin{array}{l}P_{total} = P^{\circ}_{A}x_{A} + P^{\circ}_{B}(1 – x_{A})\end{array} \)
=>
\(\begin{array}{l}P_{total} = P^{\circ}_{A}x_{A} + P^{\circ}_{B} – P^{\circ}_{B}x_{A}\end{array} \)
=>
\(\begin{array}{l}P_{total} = (P^{\circ}_{A} – P^{\circ}_{B})x_{A} + P^{\circ}_{B}\end{array} \)
=> 600 = (450 – 700) xA + 700
=> –100 = –250xA
=> xA = 0.4
Therefore,
\(\begin{array}{l}x_{B} = 1 – x_{A}\end{array} \)
= 1 – 0.4 = 0.6
Now,
\(\begin{array}{l}P_{A} = P^{\circ}_{A}x_{A}\end{array} \)
= 450 x 0.4 = 180 mm of Hg
\(\begin{array}{l}P_{B} = P^{\circ}_{B}x_{B}\end{array} \)
= 700 x 0.6 = 420 mm of Hg
Now, in the vapour phase, the mole fraction of liquid A =
\(\begin{array}{l}\frac{P_{A}}{P_{A} + P_{B}}\end{array} \)
=
\(\begin{array}{l}\frac{180}{180 + 420}\end{array} \)
=
\(\begin{array}{l}\frac{180}{600}\end{array} \)
= 0.30
And, mole fraction of liquid B = 1 – 0.30 = 0.70
Q 2.9) Find the vapour pressure of water and its relative lowering in the solution which is 50 g of urea (NH2CONH2) dissolved in 850 g of water (Vapor pressure of pure water at 298 K is 23.8 mm Hg).
Answer 2.9:
It is given that vapour pressure of water,
\(\begin{array}{l}P^{\circ}_{1}\end{array} \)
= 23.8 mm of Hg
Weight of water taken,
\(\begin{array}{l}w_{1}\end{array} \)
= 850 g
Weight of urea taken,
\(\begin{array}{l}w_{2}\end{array} \)
= 50 g
Molecular weight of water,
\(\begin{array}{l}M_{1}\end{array} \)
= 18 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Molecular weight of urea,
\(\begin{array}{l}M_{2}\end{array} \)
= 60 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Now, we have to calculate the vapour pressure of water in the solution. We take vapour pressure as p1.
Now, from Raoult’s law, we have
\(\begin{array}{l}\frac{P^{\circ}_{1} – P_{1}}{P^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}\end{array} \)
=>
\(\begin{array}{l}\frac{P^{\circ}_{1} – P_{1}}{P^{\circ}_{1}} = \frac{\frac{w_{2}}{M_{2}}}{\frac{w_{1}}{M_{1}} + \frac{w_{2}}{M_{2}}}\end{array} \)
=>
\(\begin{array}{l}\frac{23.8 – P_{1}}{23.8} = \frac{\frac{50}{60}}{\frac{850}{18} + \frac{50}{60}}\end{array} \)
=>
\(\begin{array}{l}\frac{23.8 – P_{1}}{23.8} = \frac{0.83}{47.22 + 0.83}\end{array} \)
=>
\(\begin{array}{l}\frac{23.8 – P_{1}}{23.8} = 0.0173\end{array} \)
=>
\(\begin{array}{l}P_{1}\end{array} \)
= 23.4 mm of Hg
Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg, and its relative lowering is 0.0173.
Q 2.10) How much of sucrose is to be added to 500 g of water such that it boils at 100°C if the molar elevation constant for water is 0.52 K kg mol-1 and the boiling point of water at 750 mm Hg is 99.63°C?
Answer 2.10:
Here, elevation of boiling point
\(\begin{array}{l}\Delta T_{b}\end{array} \)
= (100 + 273) – (99.63 + 273)
= 0.37 K
Mass of water,
\(\begin{array}{l}w_{1}\end{array} \)
= 500 g
Molar mass of sucrose (C12H22O11),
\(\begin{array}{l}M_{2}\end{array} \)
= 11 x 12 + 22 x 1 + 11 x 16
= 342 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Molar elevation constant, Kb = 0.52 K kg
\(\begin{array}{l}mol^{-1}\end{array} \)
We know that
\(\begin{array}{l}\Delta T_{b} = \frac{K_{b}\times 1000\times w_{2}}{M_{2}\times w_{1}}\end{array} \)
=>
\(\begin{array}{l}w_{2} = \frac{\Delta T_{b}\times M_{2}\times w_{1}}{K_{b}\times 1000}\end{array} \)
=
\(\begin{array}{l}\frac{0.37\times 342\times 500}{0.52\times 1000}\end{array} \)
= 121.67 g (approximately)
Hence, the amount of sucrose that is to be added is 121.67 g.
Q 2.11) To lower the melting point of 75 g of acetic acid by 1.50C, how much mass of ascorbic acid is needed to be dissolved in the solution where Kt = 3.9 K kg
\(\begin{array}{l}mol^{-1}\end{array} \)
?
Answer 2.11:
Mass of acetic acid (w1) = 75 g
Molar mass of ascorbic acid (C6H8O6), M2 = 6 x 12 + 8 x 1 + 6 x 16 = 176 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Lowering the melting point
\(\begin{array}{l}\Delta T_{f}\end{array} \)
= 1.5 K
We know that
\(\begin{array}{l}\Delta T_{f} = \frac{K_{f}\times w_{2}\times 1000}{M_{2}\times w_{1}}\end{array} \)
=>
\(\begin{array}{l}w_{2} = \frac{\Delta T_{f}\times M_{2}\times w_{1}}{K_{f}\times 1000}\end{array} \)
=
\(\begin{array}{l}\frac{1.5\times 176\times 75}{3.9\times 1000}\end{array} \)
= 5.08 g (approx)
Hence, the amount of ascorbic acid needed to be dissolved is 5.08 g.
Q 2.12) If a solution is prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C, calculate the osmotic pressure in Pascal exerted by it.
Answer 2.12:
It is given that
Volume of water (V) = 450 mL = 0.45 L
Temperature (T) = 37 + 273 = 310 K
Number of moles of the polymer, n =
\(\begin{array}{l}\frac{1}{185000}\end{array} \)
mol
We know that
Osmotic pressure,
\(\begin{array}{l}\pi = \frac{n}{V}\; RT\end{array} \)
=
\(\begin{array}{l}\frac{1}{185000}\;mol\times \frac{1}{0.45\;L}\times 8.314\times 10^{3}\; PaL\; K^{-1}mol^{-1}\times 310\; K\end{array} \)
= 30.98 Pa
= 31 Pa (approx)
Q 2.13) The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas?
Answer 2.13:
Molar mass of ethane (C2H6) = 2 x 12 + 6 x 1 = 30 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Therefore, number of moles present in
\(\begin{array}{l}6.56\times 10^{-2}\;g\end{array} \)
of ethane =
\(\begin{array}{l}\frac{6.56\times 10^{-2}}{30}\end{array} \)
=
\(\begin{array}{l}2.187\times 10^{-3}\;mol\end{array} \)
Let ‘x’ be the number of moles of the solvent, according to Henry’s law,
\(\begin{array}{l}p = K_{H}x\end{array} \)
=> 1 bar =
\(\begin{array}{l}K_{H}. \frac{2.187\times 10^{-3}}{2.187\times 10^{-3} + x}\end{array} \)
=> 1 bar =
\(\begin{array}{l}K_{H}. \frac{2.187\times 10^{-3}}{x}\end{array} \)
=>
\(\begin{array}{l}K_{H} = \frac{x}{2.187\times 10^{-3}}\end{array} \)
bar (Since x >>
\(\begin{array}{l}2.187\times 10^{-3}\end{array} \)
)
The number of moles present in
\(\begin{array}{l}5\times 10^{-2}\end{array} \)
g of ethane =
\(\begin{array}{l}\frac{5\times 10^{-2}}{30}\end{array} \)
mol
=
\(\begin{array}{l}1.67\times 10^{-3}\;mol\end{array} \)
According to Henry’s law,
\(\begin{array}{l}p = K_{H}x\end{array} \)
=
\(\begin{array}{l}\frac{x}{2.187\times 10^{-3}}\times \frac{1.67\times 10^{-3}}{(1.67\times 10^{-3}) + x}\end{array} \)
=
\(\begin{array}{l}\frac{x}{2.187\times 10^{-3}}\times \frac{1.67\times 10^{-3}}{x}\end{array} \)
(Since, x>>
\(\begin{array}{l}1.67\times 10^{-3}\end{array} \)
)
= 0.764 bar
Hence, the partial pressure of the gas shall be 0.764 bar.
Q 2.14) What is meant by positive and negative deviations from Raoult’s law, and how is the sign of
\(\begin{array}{l}\Delta_{sol} H\end{array} \)
related to positive and negative deviations from Raoult’s law?
Answer 2.14:
According to Raoult’s law, the partial vapour pressure of each volatile component in any solution is directly proportional to its mole fraction. The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The solutions that do not obey Raoult’s law (non-ideal solutions) have vapour pressures, either higher or lower than that predicted by Raoult’s law. If the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit negative deviation from Raoult’s law.
Vapour pressure of a two-component solution showing positive deviation from Raoult’s law
Vapour pressure of a two-component solution showing negative deviation from Raoult’s law.
In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero.
\(\begin{array}{l}\Delta _{sol}H = 0\end{array} \)
In the case of solutions showing positive deviations, the absorption of heat takes place.
\(\begin{array}{l}∴ \Delta _{sol}H = Positive\end{array} \)
In the case of solutions showing negative deviations, the evolution of heat takes place.
\(\begin{array}{l}∴ \Delta _{sol}H = Negative\end{array} \)
Q 2.15) An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Answer 2.15:
Vapour pressure of the solution at normal boiling point,
\(\begin{array}{l}p_{1}\end{array} \)
= 1.004 bar
Vapour pressure of pure water at normal boiling point,
\(\begin{array}{l}p^{\circ}_{1}\end{array} \)
= 1.013 bar
Mass of solute, w2 = 2 g
Mass of solvent (water), M1 = 18 g
\(\begin{array}{l}mol^{-1}\end{array} \)
According to Raoult’s law,
\(\begin{array}{l}\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times M_{1}}{M_{2}\times w_{1}}\end{array} \)
=>
\(\begin{array}{l}\frac{1.013 – 1.004}{1.013} = \frac{2\times 18}{M_{2}\times 98}\end{array} \)
=>
\(\begin{array}{l}\frac{0.009}{1.013} = \frac{2\times 18}{M_{2}\times 98}\end{array} \)
=>
\(\begin{array}{l}M_{2} = \frac{1.013\times 2\times 18}{0.009\times 98}\end{array} \)
= 41.35 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Hence, 41.35 g
\(\begin{array}{l}mol^{-1}\end{array} \)
is the molar mass of the solute.
Q 2.16) Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa, respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Answer 2.16:
Vapour pressure of heptanes,
\(\begin{array}{l}p^{\circ}_{1}\end{array} \)
= 105.2 kPa
Vapour pressure of octane,
\(\begin{array}{l}p^{\circ}_{2}\end{array} \)
= 46.8 kPa
We know that,
The molar mass of heptanes (C7H16) = 7 x 12 + 16 x 1 = 100 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Therefore, the number of moles of heptane =
\(\begin{array}{l}\frac{26}{100}\end{array} \)
= 0.26 mol
The molar mass of octane (C8H18) = 8 x 12 + 18 x 1 = 114 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Therefore, the number of moles of octane =
\(\begin{array}{l}\frac{35}{114}\end{array} \)
= 0.31 mol
The mole fraction of heptane,
\(\begin{array}{l}x_{1} = \frac{0.26}{0.26 + 0.31}\end{array} \)
= 0.456
And, the mole fraction of octane,
\(\begin{array}{l}x_{2} = 1 – 0.456\end{array} \)
= 0.544
Now, the partial pressure of heptane,
\(\begin{array}{l}p_{1} = x_{1}p^{\circ}_{1}\end{array} \)
= 0.456 x 105.2
= 47.97 kPa
Partial pressure of octane,
\(\begin{array}{l}p_{2} = x_{2}p^{\circ}_{2}\end{array} \)
= 0.544 x 46.8
= 25.46 kPa
Hence, vapour pressure of solution,
\(\begin{array}{l}p_{total} = p_{1} + p_{2}\end{array} \)
= 47.97 + 25.46
= 73.43 kPa
Q 2.17) The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of 1 molal solution of a non-volatile solute in it.
Answer 2.17:
1 molal solution means 1 mol of the solute is present in 100 g of the solvent (water).
The molar mass of water = 18 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Therefore, the number of moles present in 1000 g of water =
\(\begin{array}{l}\frac{1000}{18}\end{array} \)
= 55.56 mol
Therefore, the mole fraction of the solute in the solution is
\(\begin{array}{l}x_{2} = \frac{1}{1 + 55.56}\end{array} \)
= 0.0177
It is given that,
Vapour pressure of water,
\(\begin{array}{l}p^{\circ}_{1}\end{array} \)
= 12.3 kPa
Applying the relation,
\(\begin{array}{l}\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = x_{2}\end{array} \)
=>
\(\begin{array}{l}\frac{12.3 – p_{1}}{12.3}\end{array} \)
= 0.0177
=> 12.3 – p1 = 0.2177
=> p1 = 12.0823
= 12.08 kPa (approx)
Hence, the vapour pressure of the solution is 12.08 kPa.
Q 2.18) Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Answer 2.18:
Let
\(\begin{array}{l}p^{\circ}_{1}\end{array} \)
be the vapour pressure of pure octane.
Then, after dissolving the non-volatile solute, the vapour pressure of octane is
\(\begin{array}{l}\frac{80}{100}\; p^{\circ}_{1} = 0.8\; p^{\circ}_{1}\end{array} \)
The molar mass of solute, M2 = 40 g
\(\begin{array}{l}mol^{-1}\end{array} \)
The mass of octane, w1 = 114 g
The molar mass of octane, (C8H18), M1 = 8 x 12 + 18 x 1 = 114 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Applying the relation,
\(\begin{array}{l}\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times M_{1}}{M_{2}\times w_{1}}\end{array} \)
=>
\(\begin{array}{l}\frac{p^{\circ}_{1} – 0.8p^{\circ}_{1}}{p^{\circ}_{1}} = \frac{w_{2}\times 114}{40\times 114}\end{array} \)
=>
\(\begin{array}{l}\frac{0.2\; p^{\circ}_{1}}{p^{\circ}_{1}} = \frac{w_{2}}{40}\end{array} \)
=> 0.2 =
\(\begin{array}{l}\frac{w_{2}}{40}\end{array} \)
=> w2 = 8 g
Hence, the required mass of the solute is 8 g.
Q 2.19) A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution, and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K.
Answer 2.19:
(i) Let, the molar mass of the solute be M g
\(\begin{array}{l}mol^{-1}\end{array} \)
Now, the number of moles of solvent (water),
\(\begin{array}{l}n_{1} = \frac{90\; g}{18\; g\; mol^{-1}}\end{array} \)
= 5 mol
And, the number of moles of solute,
\(\begin{array}{l}n_{2} = \frac{30\; g}{M\; mol^{-1}} = \frac{30}{M}\; mol\end{array} \)
p1 = 2.8 kPa
Applying the relation:
\(\begin{array}{l}\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}\end{array} \)
=>
\(\begin{array}{l}\frac{p^{\circ}_{1} – 2.8}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{5 + \frac{30}{M}}\end{array} \)
=>
\(\begin{array}{l}1 – \frac{2.8}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{\frac{5M + 30}{M}}\end{array} \)
=>
\(\begin{array}{l}1 – \frac{2.8}{p^{\circ}_{1}} = \frac{30}{5M + 30}\end{array} \)
=>
\(\begin{array}{l}\frac{2.8}{p^{\circ}_{1}} = 1 – \frac{30}{5M + 30}\end{array} \)
=>
\(\begin{array}{l}\frac{2.8}{p^{\circ}_{1}} = \frac{5M + 30 – 30}{5M + 30}\end{array} \)
=>
\(\begin{array}{l}\frac{2.8}{p^{\circ}_{1}} = \frac{5M}{5M + 30}\end{array} \)
=>
\(\begin{array}{l}\frac{p^{\circ}_{1}}{2.8} = \frac{5M + 30}{5M}\end{array} \)
(i)
After the addition of 18 g of water:
\(\begin{array}{l}n_{1} = \frac{90 + 18g}{18} = 6\;mol\end{array} \)
\(\begin{array}{l}p_{1} = 2.9 \;kPa\end{array} \)
Again applying the relation:
\(\begin{array}{l}\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{2}}{n_{1} + n_{2}}\end{array} \)
=>
\(\begin{array}{l}\frac{p^{\circ}_{1} – 2.9}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{6 + \frac{30}{M}}\end{array} \)
=>
\(\begin{array}{l}1 – \frac{2.9}{p^{\circ}_{1}} = \frac{\frac{30}{M}}{\frac{6M + 30}{M}}\end{array} \)
=>
\(\begin{array}{l}1 – \frac{2.9}{p^{\circ}_{1}} = \frac{30}{6M + 30}\end{array} \)
=>
\(\begin{array}{l}\frac{2.9}{p^{\circ}_{1}} = 1 – \frac{30}{6M + 30}\end{array} \)
=>
\(\begin{array}{l}\frac{2.9}{p^{\circ}_{1}} = \frac{6M + 30 – 30}{6M + 30}\end{array} \)
=>
\(\begin{array}{l}\frac{2.9}{p^{\circ}_{1}} = \frac{6M}{6M + 30}\end{array} \)
=>
\(\begin{array}{l}\frac{p^{\circ}_{1}}{2.9} = \frac{6M + 30}{6M}\end{array} \)
(ii)
Dividing equation (i) by (ii), we have
\(\begin{array}{l}\frac{2.9}{2.8} = \frac{\frac{5M + 30}{5M}}{\frac{6M + 30}{6M}}\end{array} \)
=>
\(\begin{array}{l}\frac{2.9}{2.8}\times \frac{6M + 30}{6} = \frac{5M + 30}{5}\end{array} \)
=>
\(\begin{array}{l}2.9\times 5\times (6M + 30) = 2.8\times 6\times (5M + 30)\end{array} \)
=> 87M + 435 = 84M + 504
=> 3M = 69
=> M = 23 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Therefore, 23 g
\(\begin{array}{l}mol^{-1}\end{array} \)
is the molar mass of the solute.
(ii) Putting the value of ‘M’ in equation (i), we have
\(\begin{array}{l}\frac{p^{\circ}_{1}}{2.8} = \frac{5\times 23 + 30}{5\times 23}\end{array} \)
=>
\(\begin{array}{l}\frac{p^{\circ}_{1}}{2.8} = \frac{145}{115}\end{array} \)
=>
\(\begin{array}{l}p^{\circ}_{1}\end{array} \)
= 3.53 kPa
Hence, 3.53 kPa is the vapour pressure of water at 298 K.
Q 2.20) A 5% solution (by mass) of cane sugar in water has a freezing point of 271K. Calculate the freezing point of 5% glucose in water if the freezing point of pure water is 273.15 K.
Answer 2.20:
\(\begin{array}{l}\Delta T_{f}\end{array} \)
= 273.15 – 271 = 2.15 K
The molar mass of sugar (C12H22O11) = 12 x 12 + 22 x 1 + 11 x 16 = 342 g
\(\begin{array}{l}mol^{-1}\end{array} \)
5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 − 5)g
= 95 g of water.
Now, the number of moles of cane sugar =
\(\begin{array}{l}\frac{5}{342}\end{array} \)
mol = 0.0146 mol
Therefore, the molality of the solution,
\(\begin{array}{l}m = \frac{0.0146\; mol}{0.095\; kg} = 0.1537 \;mol\;kg^{-1}\end{array} \)
Applying the relation,
\(\begin{array}{l}\Delta T_{f} = K_{f}\times m\end{array} \)
=>
\(\begin{array}{l}K_{f} = \frac{\Delta T_{f}}{m}\end{array} \)
=
\(\begin{array}{l}\frac{2.15\; K}{0.1537\;mol\;kg^{-1}}\end{array} \)
= 13.99 K kg
\(\begin{array}{l}mol^{-1}\end{array} \)
The molar mass of glucose (C6H12O6) = 6 x 12 + 12 x 1 + 6 x 16 = 180 g
\(\begin{array}{l}mol^{-1}\end{array} \)
5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water.
Therefore, the number of moles of glucose =
\(\begin{array}{l}\frac{5}{180}\end{array} \)
mol = 0.0278 mol
Therefore, the molality of the solution, m =
\(\begin{array}{l}\frac{0.0278\;mol}{0.095\;kg}\end{array} \)
= 0.2926 mol
\(\begin{array}{l}kg^{-1}\end{array} \)
Applying the relation:
\(\begin{array}{l}\Delta T_{f} = K_{f}\times m\end{array} \)
=
\(\begin{array}{l}13.99\;K\;kg\;mol^{-1}\times 0.2926\;mol\;kg^{-1}\end{array} \)
= 4.09 K (approx)
Hence, the freezing point of the 5 % glucose solution is (273.15 – 4.09) K = 269.06 K.
Q 2.21) Two elements A and B form compounds having formulas AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K, whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol–1. Calculate the atomic masses of A and B.
Answer 2.21:
We know that,
\(\begin{array}{l}M_{2} = \frac{1000\times w_{2}\times k_{f}}{\Delta T_{f}\times w_{1}}\end{array} \)
Then,
\(\begin{array}{l}M_{AB_{2}} = \frac{1000\times 1\times 5.1}{2.3\times 20}\end{array} \)
= 110.87 g
\(\begin{array}{l}mol^{-1}\end{array} \)
\(\begin{array}{l}M_{AB_{4}} = \frac{1000\times 1\times 5.1}{1.3\times 20}\end{array} \)
= 196.15 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Now, we have the molar masses of AB2 and AB4 as 110.87 g
\(\begin{array}{l}mol^{-1}\end{array} \)
and 196.15 g
\(\begin{array}{l}mol^{-1}\end{array} \)
respectively.
Let the atomic masses of A and B be x and y, respectively.
Now, we can write:
x + 2y = 110.87 …(i)
x + 4y = 196.15 …(ii)
Subtracting equation (i) from (ii), we have
2y = 85.28
=> y = 42.64
Putting the value of ‘y’ in equation (1), we have
x + 2 (42.64) = 110.87
=> x = 25.59
Hence, the atomic masses of A and B are 25.59 u and 42.64 u, respectively.
Q 2.22) At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Answer 2.22:
Given:
T = 300 K
n = 1.52 bar
R = 0.083 bar L
\(\begin{array}{l}K^{-1}\; mol^{-1}\end{array} \)
Applying the relation, n = CRT
=> C =
\(\begin{array}{l}\frac{n}{RT}\end{array} \)
=
\(\begin{array}{l}\frac{1.52\;bar}{0.083\;bar\;L\;K^{-1}\;mol^{-1}\times 300\;K}\end{array} \)
= 0.061 mol
Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.
Q 2.23) Suggest the most important type of intermolecular attractive interaction in
the following pairs.
(i) n-hexane and n-octane
(ii) I2 and CCl4
(iii) NaClO4 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H6O)
Answer 2.23:
(i) Van der Wall’s forces of attraction
(ii) Van der Wall’s forces of attraction
(iii) Ion-dipole interaction
(iv) Dipole-dipole interaction
(v) Dipole-dipole interaction
Q 2.24) Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.
Answer 2.24:
n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in the n-octane.
The order of increasing polarity is
Cyclohexane < CH3CN < CH3OH < KCl
Therefore, the order of increasing solubility is
KCl < CH3OH < CH3CN < Cyclohexane
Q 2.25) Amongst the following compounds, identify which are insoluble, partially
soluble and highly soluble in water?
(i) phenol (ii) toluene (iii) formic acid
(iv) ethylene glycol (v) chloroform (vi) pentanol
Answer 2.25:
(i) Phenol (C6H5OH) has the polar group −OH and non-polar group –C6H5. Thus, phenol is partially soluble in water.
(ii) Toluene (C6H5−CH3) has no polar groups. Thus, toluene is insoluble in water.
(iii) Formic acid (HCOOH) has the polar group −OH and can form H-bond with water.
Thus, formic acid is highly soluble in water.
(iv) Ethylene glycol has a polar −OH group and can form H−bond. Thus, it is highly soluble in water.
(v) Chloroform is insoluble in water.
(vi) Pentanol (C5H11OH) has a polar −OH group, but it also contains a very bulky nonpolar −C5H11 group. Thus, pentanol is partially soluble in water.
Q 2.26) If the density of some lake water is 1.25g mL–1 and contains 92 g of Na+ ions per kg of water, calculate the molarity of Na+ ions in the lake.
Answer 2.26:
The number of moles present in 92 g of Na+ ions =
\(\begin{array}{l}\frac{92\;g}{23\;g\;mol^{-1}}\end{array} \)
= 4 mol
Therefore, the molality of Na+ ions in the lake =
\(\begin{array}{l}\frac{4\;mol}{1\;kg}\end{array} \)
= 4 m
Q 2.27) If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of
CuS in aqueous solution.
Answer 2.27:
Solubility product of CuS,
\(\begin{array}{l}K_{sp} = 6\times 10^{-16}\end{array} \)
Let s be the solubility of CuS in mol L-1.
\(\begin{array}{l}CuS \leftrightarrow Cu^{2+} + S^{2-}\end{array} \)
Now, s s
\(\begin{array}{l}K_{sp} = [Cu^{2+}] + [S^{2-}]\end{array} \)
= s x s
= s2
Then, we have,
\(\begin{array}{l}K_{sp} = s^{2} = 6\times 10^{-16}\end{array} \)
=>
\(\begin{array}{l}s = \sqrt{6\times 10^{-16}}\end{array} \)
=
\(\begin{array}{l}2.45\times 10^{-8}\;mol\;L^{-1}\end{array} \)
Hence,
\(\begin{array}{l}2.45\times 10^{-8}\;mol\;L^{-1}\end{array} \)
is the maximum molarity of CuS in an aqueous solution.
Q 2.28) Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when
6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.
Answer 2.28:
6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.
Then, the total mass of the solution = (6.5 + 450) g = 456.5 g
Therefore, the mass percentage of C9H8O4 =
\(\begin{array}{l}\frac{6.5}{456.5}\times 100\end{array} \)
= 1.424%
Q 2.29) Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. The dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10–3 m aqueous solution required for the above dose.
Answer 2.29:
The molar mass of nalorphene (C19H21NO3) = 19 x 12 + 21 x 1 + 1 x 14 + 3 x 16 = 311 g mol-1
In
\(\begin{array}{l}1.5\times 10^{-3}\;m\end{array} \)
aqueous solution of nalorphene,
1 kg (1000 g) of water contains
\(\begin{array}{l}1.5\times 10^{-3}\;mol\end{array} \)
=
\(\begin{array}{l}1.5\times 10^{-3}\times 311\end{array} \)
g
= 0.4665 g
Therefore, total mass of the solution = (1000 + 0.4665) g = 1000.4665 g
This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665 g.
Therefore, the mass of the solution containing 1.5 mg of nalorphene is
\(\begin{array}{l}\frac{1000.4665\times 1.5\times 10^{-3}}{0.4665}\end{array} \)
g
= 3.22 g
Hence, 3.22 g is the required mass of the aqueous solution.
Q 2.30) Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250
mL of 0.15 M solutions in methanol.
Answer 2.30:
0.15 M solution of benzoic acid in methanol means,
1000 mL of solution contains 0.15 mol of benzoic acid.
Therefore, 250 mL of solution contains
\(\begin{array}{l}\frac{0.15\times 250}{1000}\end{array} \)
mol of benzoic acid
= 0.0375 mol of benzoic acid
Molar mass of benzoic acid (C6H5COOH) = 7 x 12 + 6 x 1 + 2 x 16 = 122 g mol-1
Hence, required benzoic acid = 0.0375 mol x 122 g mol-1 = 4.575 g
Q 2.31) The depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
Answer 2.31:
Among H, Cl, and F, H is the least electronegative, while F is the most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H+ ions, i.e., trifluoroacetic acid ionises to the largest extent. Now, the more ions produced, the greater the depression of the freezing point. Hence, the depression in the freezing point increases in the order
Acetic acid < trichloroacetic acid < trifluoroacetic acid
Q 2.32) Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10–3, Kf = 1.86 K kg mol–1
Answer 2.32:
Molar mass of CH3CH2CHCICOOH = 15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1
= 122.5 g mol-1
Therefore, the number of moles present in 10 g of CH3CH2CHCICOOH =
\(\begin{array}{l}\frac{10\;g}{122.5\;g\;mol^{-1}}\end{array} \)
= 0.0816 mol
It is given that 10 g of CH3CH2CHCICOOH is added to 250 g of water.
Therefore, the molality of the solution, CH3CH2CHCICOOH =
\(\begin{array}{l}\frac{0.0186}{250}\times 1000\end{array} \)
= 0.3264 mol kg-1
Let ‘a’ be the degree of dissociation of CH3CH2CHCICOOH.
CH3CH2CHCICOOH undergoes dissociation according to the following equation:
\(\begin{array}{l}∴ K_{a} = \frac{C\alpha .C\alpha}{C (1 – \alpha)}\end{array} \)
=
\(\begin{array}{l}\frac{C\alpha^{2}}{1 – \alpha}\end{array} \)
Since a is very small with respect to 1,
\(\begin{array}{l}1 – a \approx 1\end{array} \)
\(\begin{array}{l}K_{a} = \frac{C \alpha^{2}}{1}\end{array} \)
Now,
=>
\(\begin{array}{l}K_{a} = C \alpha^{2}\end{array} \)
=>
\(\begin{array}{l}\alpha = \sqrt{\frac{K_{a}}{C}}\end{array} \)
=
\(\begin{array}{l}\sqrt{\frac{1.4\times 10^{-3}}{0.3264}}\;\;\;\;(∵ K_{a}=1.4\times 10^{-3})\end{array} \)
= 0.0655
Again,
Total moles of equilibrium = 1 – a + a + a = 1 + a
\(\begin{array}{l}∴ i = \frac{1 + \alpha}{1}\end{array} \)
=
\(\begin{array}{l}1 + \alpha\end{array} \)
= 1 + 0.0655
= 1.0655
Hence, the depression in the freezing point of water is given as
\(\begin{array}{l}\Delta T_{f} = i.K_{f}m\end{array} \)
=
\(\begin{array}{l}1.0655\times 1.86\;K\;kg\;mol^{-1}\times 0.3264\;mol\;kg^{-1}\end{array} \)
= 0.65 K
Q 2.33) 19.5 g of CH2 FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
Answer 2.33:
Given:
w1 = 500 g
w2 = 19.5 g
Kf = 1.86 K kg
\(\begin{array}{l}mol^{-1}\end{array} \)
\(\begin{array}{l}\Delta T_{f}\end{array} \)
= 1 K
We know that
\(\begin{array}{l}M_{2} = \frac{K_{f}\times w_{2}\times 1000}{\Delta T_{f}\times w_{1}}\end{array} \)
=
\(\begin{array}{l}\frac{1.86\;K\;kg\;mol^{-1}\times 19.5\;g\times 1000\;g\;kg^{-1}}{500\;g\times 1\;K}\end{array} \)
=
\(\begin{array}{l}72.54\;g\;mol^{-1}\end{array} \)
Therefore, observed molar mass of CH2FCOOH,
\(\begin{array}{l}(M_{2})_{obs} = 72.54\;g\;mol^{-1}\end{array} \)
The calculated molar mass of CH2FCOOH,
\(\begin{array}{l}(M_{2})_{cal}\end{array} \)
= 14 + 19 + 12 + 16 + 16 + 1 = 78 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Therefore, van’t Hoff factor,
\(\begin{array}{l}i = \frac{(M_{2})_{cal}}{(M_{2})_{obs}}\end{array} \)
is:
=
\(\begin{array}{l}\frac{78\;g\;mol^{-1}}{72.54\;g\;mol^{-1}}\end{array} \)
= 1.0753
Let ‘a’ be the degree of dissociation of CH2FCOOH.
\(\begin{array}{l}∴ i = \frac{C(1 + \alpha)}{C}\end{array} \)
=> i = 1 +
\(\begin{array}{l}\alpha\end{array} \)
=>
\(\begin{array}{l}\alpha\end{array} \)
= i – 1
= 1.0753 – 1
= 0.0753
Now, the value of Ka is given as
\(\begin{array}{l}K_{a} = \frac{[CH_{2}FCOO^{-}][H^{+}]}{[CH_{2}FCOOH]}\end{array} \)
=
\(\begin{array}{l}\frac{C\alpha.\; C\alpha}{C(1 – \alpha)}\end{array} \)
=
\(\begin{array}{l}\frac{C\alpha^{2}}{1 – \alpha}\end{array} \)
Taking the volume of the solution as 500 mL, we have the concentration
C =
\(\begin{array}{l}\frac{\frac{19.5}{78}}{500}\times 1000\;M\end{array} \)
= 0.5 M
Therefore,
\(\begin{array}{l}K_{a} = \frac{C\alpha^{2}}{1 – \alpha}\end{array} \)
=
\(\begin{array}{l}\frac{0.5\times (0.0753)^{2}}{1 – 0.0753}\end{array} \)
=
\(\begin{array}{l}\frac{0.5\times 0.00567}{0.9247}\end{array} \)
= 0.00307 (approx)
=
\(\begin{array}{l}3.07\times 10^{-3}\end{array} \)
Q 2.34) Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Answer 2.34:
Vapour pressure of water,
\(\begin{array}{l}p^{\circ}_{1}\end{array} \)
= 17.535 mm of Hg
Mass of glucose, w2 = 25 g
Mass of water, w1 = 450 g
We know that,
Molar mass of glucose (C6H12O6), M2 = 6 x 12 + 12 x 1 + 6 x 16 = 180 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Molar mass of water, M1 = 18 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Then, the number of moles of glucose,
\(\begin{array}{l}n_{2} = \frac{25}{180\;g\;mol^{-1}}\end{array} \)
= 0.139 mol
And, the number of moles of water,
\(\begin{array}{l}n_{1} = \frac{450\;g}{18\;g\;mol^{-1}}\end{array} \)
= 25 mol
We know that,
\(\begin{array}{l}\frac{p^{\circ}_{1} – p_{1}}{p^{\circ}_{1}} = \frac{n_{1}}{n_{2} + n_{1}}\end{array} \)
=>
\(\begin{array}{l}\frac{17.535 – p_{1}}{17.535} = \frac{0.139}{0.139 + 25}\end{array} \)
=>
\(\begin{array}{l}17.535 – p_{1} = \frac{0.139\times 17.535}{25.139}\end{array} \)
=>
\(\begin{array}{l}17.535 – p_{1} = 0.097\end{array} \)
=>
\(\begin{array}{l}p_{1} = 17.44\end{array} \)
mm of Hg
Hence, 17.44 mm of Hg is the vapour pressure of water.
Q 2.35) Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Answer 2.35:
Given:
p = 760 mm Hg
\(\begin{array}{l}k_{H} = 4.27\times 10^{5}\end{array} \)
mm Hg
According to Henry’s law,
p = kHx
=> x =
\(\begin{array}{l}\frac{p}{k_{H}}\end{array} \)
=
\(\begin{array}{l}\frac{760\;mm\;Hg}{4.27\times 10^{5}\;mm\;Hg}\end{array} \)
=
\(\begin{array}{l}177.99\times 10^{-5}\end{array} \)
=
\(\begin{array}{l}178\times 10^{-5}\end{array} \)
(approx)
Hence,
\(\begin{array}{l}178\times 10^{-5}\end{array} \)
is the mole fraction of methane in benzene.
Q 2.36) 100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour pressure of pure liquid B was found to be 500 torrs. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Answer 2.36:
Number of moles of liquid A,
\(\begin{array}{l}n_{A} = \frac{100}{140}\end{array} \)
= 0.714 mol
Number of moles of liquid B,
\(\begin{array}{l}n_{B} = \frac{1000}{180}\end{array} \)
= 5.556 mol
Then, the mole fraction of A,
\(\begin{array}{l}x_{A} = \frac{n_{A}}{n_{A} + n_{B}}\end{array} \)
=
\(\begin{array}{l}\frac{0.714}{0.714 + 5.556}\end{array} \)
= 0.114
And, the mole fraction of B, xB = 1 – 0.114 = 0.886
Vapour pressure of pure liquid B,
\(\begin{array}{l}p^{\circ}_{B}\end{array} \)
= 500 torr
Therefore, the vapour pressure of liquid B in the solution,
\(\begin{array}{l}p_{B} = p^{\circ}_{B}\;x_{B}\end{array} \)
= 500 x 0.886
= 443 torr
Total vapour pressure of the solution,
\(\begin{array}{l}p_{total}\end{array} \)
= 475 torr
Therefore, the vapour pressure of liquid A in the solution,
\(\begin{array}{l}p_{A} = p_{total} – p_{B}\end{array} \)
= 475 – 443
= 32 torr
Now,
\(\begin{array}{l}p_{A} = p^{\circ}_{A}\;x_{A}\end{array} \)
=>
\(\begin{array}{l}p^{\circ}_{A} = \frac{p_{A}}{x_{A}}\end{array} \)
=
\(\begin{array}{l}\frac{32}{0.114}\end{array} \)
= 280.7 torr
Hence, 280.7 torr is the vapour pressure of pure liquid A.
Q 2.37) Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form the ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetone as a function of xacetone.The experimental data observed for different compositions of the mixture is:
| \(\begin{array}{l}100\times x_{acetone}\end{array} \) | 0 | 11.8 | 23.4 | 36.0 | 50.8 | 58.2 | 64.5 | 72.1 |
| \(\begin{array}{l}p_{acetone}\end{array} \) / mm Hg | 0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
| \(\begin{array}{l}p_{chloroform}\end{array} \) / mm Hg | 632.8 | 548.1 | 469.4 | 359.7 | 257.7 | 193.6 | 161.2 | 120.7 |
Plot this data also on the same graph paper. Indicate whether it has a positive deviation or a negative deviation from the ideal solution.
Answer 2.37:
From the question, we have the following data
| \(\begin{array}{l}100\times x_{acetone}\end{array} \) | 0 | 11.8 | 23.4 | 36.0 | 50.8 | 58.2 | 64.5 | 72.1 |
| \(\begin{array}{l}p_{acetone}\end{array} \) / mm Hg | 0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
| \(\begin{array}{l}p_{chloroform}\end{array} \) / mm Hg | 632.8 | 548.1 | 469.4 | 359.7 | 257.7 | 193.6 | 161.2 | 120.7 |
| \(\begin{array}{l}p_{total}\end{array} \) (mm Hg) | 632.8 | 603.0 | 579.5 | 562.1 | 580.4 | 599.5 | 615.3 | 641.8 |
It can be observed from the graph that the plot for the
\(\begin{array}{l}p_{total}\end{array} \)
of the solution curves downwards. Therefore, the solution shows a negative deviation from the ideal behaviour.
Q 2.38) Benzene and toluene form the ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg, respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Answer 2.38:
Molar mass of benzene (C6H6) = 6 x 12 + 6 x 1 = 78 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Molar mass of toluene (C6H5CH3) = 7 x 12 + 8 x 1 = 92 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Now, the number of moles present in 80 g of benzene =
\(\begin{array}{l}\frac{80}{78}\end{array} \)
= 1.026 mol
And, the number of moles present in 100 g of toluene =
\(\begin{array}{l}\frac{100}{92}\end{array} \)
= 1.087 mol
Therefore, the mole fraction of benzene,
\(\begin{array}{l}x_{b} = \frac{1.026}{1.026 + 1.087}\end{array} \)
= 0.486
And, the mole fraction of toluene,
\(\begin{array}{l}x_{t}\end{array} \)
= 1 – 0.486 = 0.514
It is given that the vapour pressure of pure benzene,
\(\begin{array}{l}p^{\circ}_{b}\end{array} \)
= 50.71 mm Hg
And, the vapour pressure of pure toluene,
\(\begin{array}{l}p^{\circ}_{t}\end{array} \)
= 32.06 mm Hg
Therefore, the partial pressure of benzene,
\(\begin{array}{l}p_{b} = x_{b}\times p^{\circ}_{b}\end{array} \)
= 0.486 x 50.71
= 24.645 mm Hg
And, the partial vapour pressure of toluene,
\(\begin{array}{l}p_{t} = x_{t}\times p^{\circ}_{t}\end{array} \)
= 0.514 x 32.06
= 16.479 mm Hg
Hence, the mole fraction of benzene in the vapour phase is given by
\(\begin{array}{l}\frac{p_{b}}{p_{b} + p_{t}}\end{array} \)
=
\(\begin{array}{l}\frac{24.645}{24.645 + 16.479}\end{array} \)
=
\(\begin{array}{l}\frac{24.645}{41.124}\end{array} \)
= 0.599
= 0.6 (approx)
Q 2.39) The air is a mixture of a number of gases. The major components are oxygen and nitrogen, with an approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K, if Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 107 mm and 6.51 × 107 mm, respectively, calculate the composition of these gases in water.
Answer 2.39:
Percentage of oxygen in air = 20 %
Percentage of nitrogen in air = 79 %
Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is (10 x 760) mm Hg = 7600 mm Hg
Therefore,
Partial pressure of oxygen,
\(\begin{array}{l}p_{O_{2}} = \frac{20}{100}\times 7600\end{array} \)
mm Hg
= 1520 mm Hg
Partial pressure of nitrogen,
\(\begin{array}{l}p_{N_{2}} = \frac{79}{100}\times 7600\end{array} \)
= 6004 mm Hg
For oxygen:
\(\begin{array}{l}p_{O_{2}} = K_{H}. x_{O_{2}}\end{array} \)
=>
\(\begin{array}{l}x_{O_{2}} = \frac{p_{O_{2}}}{K_{H}}\end{array} \)
=
\(\begin{array}{l}\frac{1520\;mm\;Hg}{3.30\times 10^{7}\;mm\;Hg}\;\;\;\;\;\;\;\;\;(Given\;K_{H} = 3.30\times 10^{7}\;mm\;Hg)\end{array} \)
=
\(\begin{array}{l}4.61\times 10^{-5}\end{array} \)
For nitrogen:
\(\begin{array}{l}p_{N_{2}} = K_{H}. x_{N_{2}}\end{array} \)
=>
\(\begin{array}{l}x_{N_{2}} = \frac{p_{N_{2}}}{K_{H}}\end{array} \)
=
\(\begin{array}{l}\frac{6004\;mm\;Hg}{6.51\times 10^{7}\;mm\;Hg}\end{array} \)
=
\(\begin{array}{l}9.22\times 10^{-5}\end{array} \)
Hence,
\(\begin{array}{l}4.61\times 10^{-5}\end{array} \)
and
\(\begin{array}{l}9.22\times 10^{-5}\end{array} \)
are the mole fractions of oxygen and nitrogen in the water.
Q 2.40) Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litres of water such that its osmotic pressure is 0.75 atm at 27° C.
Answer 2.40:
We know that,
\(\begin{array}{l}\pi = i\; \frac{n}{V}\; RT\end{array} \)
=>
\(\begin{array}{l}\pi = i\; \frac{w}{MV}\; RT\end{array} \)
=>
\(\begin{array}{l}w = \frac{\pi MV}{iRT}\end{array} \)
\(\begin{array}{l}\pi\end{array} \)
= 0.75 atm
V = 2.5 L
i = 2.47
T = (27 + 273) = 300 K
Here,
R =
\(\begin{array}{l}0.0821\;L\;atm\;K^{-1}\;mol^{-1}\end{array} \)
M = 1 x 40 + 2 x 35.5
= 111 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Therefore, w =
\(\begin{array}{l}\frac{0.75\times 111\times 2.5}{2.47\times 0.0821\times 300}\end{array} \)
= 3.42 g
Hence, 3.42 g is the required amount of CaCl2.
Q 2.41) Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4
in 2 litres of water at 25° C, assuming that it is completely dissociated.
Answer 2.41:
When K2SO4 is dissolved in water,
\(\begin{array}{l}K^{+}\; and\; SO_{4}^{2-}\end{array} \)
ions are produced.
\(\begin{array}{l}K_{2}SO_{4} \rightarrow 2K^{+} + SO_{4}^{2-}\end{array} \)
Total number of ions produced = 3
Therefore, i = 3
Given:
w = 25 mg = 0.025 g
V = 2 L
T = 250C = (25 + 273) = 298 K
Also, we know that
R =
\(\begin{array}{l}0.0821\;L\;atm\;K^{-1}\;mol^{-1}\end{array} \)
M = (2 x 39) + (1 x 32) + (4 x 16) = 174 g
\(\begin{array}{l}mol^{-1}\end{array} \)
Applying the following relation,
\(\begin{array}{l}\pi = i\; \frac{n}{V}\; RT\end{array} \)
=
\(\begin{array}{l}i \frac{w}{M} \frac{1}{v} RT\end{array} \)
=
\(\begin{array}{l}3\times \frac{0.025}{174}\times \frac{1}{2}\times 0.0821\times 298\end{array} \)
=
\(\begin{array}{l}5.27\times 10^{-3}\; atm\end{array} \)
A solution is a homogeneous mixture of two or more than two components. Solved answers on the chapter’s NCERT Solutions for Class 12 are given by our subject experts. These NCERT Solutions will help the students to understand the types of solutions, expressing the concentration of the solution, solubility, ideal and non-ideal solutions, colligative properties and abnormal molar mass. Chemistry Class 12 questions and solutions in Chapter 2 given here are very simple and easy to understand.
Class 12 NCERT Solutions for Chemistry Chapter 2 Solutions
Chapter 2 Solutions of Class 12 Chemistry, is designed as per the CBSE Syllabus for the session 2023-24. This chapter holds approximately 5 marks in the board examination. The Chemistry Class 12 NCERT Solutions Chapter 2 teaches about the types of solutions, the concentration of solutions, the solubility of solids and gases in a liquid, the vapour pressure of liquid solutions, Raoult’s law, ideal and non-ideal solutions, colligative properties and determination of molar masses.
Subtopics for Class 12 Chemistry Chapter 2 – Solutions
- Types of Solutions
- Expressing Concentration of Solutions
- Solubility
- The solubility of a Solid in a Liquid
- The solubility of a Gas in a Liquid
- Vapour Pressure of Liquid Solutions
- Vapour Pressure of Liquid-Liquid Solutions
- Raoult’s Law as a Special Case of Henry’s Law
- Vapour Pressure of Solutions of Solids in Liquids
- Ideal and Nonideal Solutions
- Ideal Solutions
- Non-ideal Solutions
- Colligative Properties and Determination of Molar Mass
- Relative Lowering of Vapour Pressure
- Elevation of Boiling Point
- Depression of Freezing Point
- Osmosis and Osmotic Pressure
- Abnormal Molar Masses
NCERT Chemistry book includes a chapter on solutions to introduce various important concepts to the students. In this, students learn to determine the molarity, molality and mole fraction of solutions and know about Henry’s law constant, mass percentage, etc. The topics provided in the NCERT books are not only important for the Class 12 board examination but also for competitive exams like JEE Main and NEET. JEE Main is a national-level engineering entrance examination, and NEET is a national-level examination to take admission to the best medical colleges in India. Also, students must solve NCERT exemplar problems, MCQS, and short and long-answer questions in order to attain a firm grip across subjects.
Solving the NCERT questions using NCERT Solutions will help them to understand the topics in an effective and simple way. Sometimes, the questions are also asked in the JEE Main and NEET examinations.
Students preparing for CBSE examinations are strictly advised to study the chapter thoroughly and should solve the NCERT questions provided at the end of each chapter. The NCERT Solutions are designed in such a way that students can understand each and every chapter of the book. Along with the NCERT questions, students are also encouraged to solve the CBSE Class 12 previous years’ question papers and sample papers. Solving the previous years’ questions and sample papers will help them get acquainted with the different types of questions and their marking scheme. Students are always encouraged to solve the previous years’ questions to boost exam preparations. Solving the question papers in a stipulated time of 3 hours will help them to understand their problem-solving speed and identify the time they take to solve a particular type of question.
Frequently Asked Questions on NCERT Solutions for Class 12 Chemistry Chapter 2
Q1
Which are the important topics in Chapter 2 of NCERT Solutions for Class 12 Chemistry?
The important topics in Chapter 2 of NCERT Solutions for Class 12 Chemistry are
Types of Solutions
Expressing Concentration of Solutions
Solubility
The Solubility of a Solid in a Liquid
The Solubility of a Gas in a Liquid
Vapour Pressure of Liquid Solutions
Vapour Pressure of Liquid-Liquid Solutions
Raoult’s Law as a Special Case of Henry’s Law
Vapour Pressure of Solutions of Solids in Liquids
Ideal and Non-ideal Solutions
Ideal Solutions
Non-ideal Solutions
Colligative Properties and Determination of Molar Mass
Relative Lowering of Vapour Pressure
Elevation of Boiling Point
Depression of Freezing Point
Osmosis and Osmotic Pressure
Abnormal Molar Masses
Q2
Can I get the NCERT Solutions for Class 12 Chemistry Chapter 2 PDF online?
The NCERT Solutions for Class 12 Chemistry Chapter 2 PDF is available on ANAND CLASSES (A School Of Competitions). These solutions are provided with a free download option which can be accessed by the students effortlessly. Students can download the PDF and refer to them while answering the textbook questions to get their doubts cleared instantly. The main advantage of using these solutions is they can be used by the students anywhere and at any time without any difficulty.
Q3
Do the NCERT Solutions for Class 12 Chemistry Chapter 2 help students with their board exam preparation?
The NCERT Solutions for Class 12 Chemistry Chapter 2 are very helpful for the students for various purposes. The expert faculty at ANAND CLASSES (A School Of Competitions) make use of a simple and understandable language, which makes it easy for the students while learning. These solutions provide students with a strong foundation of fundamental concepts which are important from the exam point of view. By using these solutions, students can save a lot of time in searching for the correct answer as per the latest CBSE Syllabus and its guidelines.
NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions provide accurate and adequate Chemistry knowledge. The NCERT Solutions for Class 12 Chemistry can be utilised by the students to prepare for the board examination and to solve the exercise questions of the Class 12 Chemistry Chapter 2. All the solutions are created according to the latest CBSE Syllabus for 2023-24 and its guidelines.
These NCERT Solutions for Class 12 Chemistry help students clear all their doubts related to this chapter with ease. Further, students might also use this resource for preparing notes and performing revisions. Click the link given below to download the NCERT Solutions for Class 12 Chemistry Chapter 2 PDF for free.