Comprehension:
Solutions
The solution which obeys Raoult's law over the entire range of concentration are known as ideal solutions and which does not obey are known as non-ideal solution. The vapour pressure of a non ideal solution is either higher or lower than that predicted by Raoult's law equation. The colligative properties, which depend on the number of solute particles irrespective of their nature, are relative lowering of vapour pressure, depression of freezing point, elevation of boiling point and osmotic pressure. All these properties can be utilised for the determination of molar mass of the solute.
2.00 g of a non-volatile, non-electrolytic solute when dissolved in 100 g of ethanol lowered the freezing point of ethanol by 0.33 K. The freezing point depression constant of ethanol is 1.99 K kg mol−1. The molar mass of the solute is-.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 120.6 g mol−1
The depression of freezing point of a dilute solution of a non-volatile, non-electrolyte solute is proportional to its molality:
\( \displaystyle \Delta T_f=K_f\,m \)
The molality is the number of moles of solute per kilogram of solvent. With w2 g of solute of molar mass M2 in w1 g of solvent:
\( \displaystyle m=\frac{w_2/M_2}{w_1/1000}=\frac{w_2\times1000}{M_2\times w_1} \)
\( \displaystyle \Delta T_f=\frac{K_f\times w_2\times1000}{M_2\times w_1} \)
\( \displaystyle M_2=\frac{K_f\times w_2\times1000}{\Delta T_f\times w_1} \)
Here Kf = 1.99 K kg mol−1, w2 = 2.00 g, w1 = 100 g and ΔTf = 0.33 K:
\( \displaystyle M_2=\frac{1.99\times2.00\times1000}{0.33\times100} \)
\( \displaystyle M_2=\frac{3980}{33} \)
\( \displaystyle M_2=120.6\ \mathrm{g\,mol^{-1}} \)
So the molar mass of the solute is 120.6 g mol−1.
Shortcut: 1.99 ≈ 2, so M2 ≈ (2 × 2 × 1000)/(0.33 × 100) = 4000/33 ≈ 121. Only Option (3) is close.