Question

A 0.10 M solution of a monoprotic acid (d = 1.01 g/cm³) is 5% ionized. What is the freezing point of the solution ? The molecular weight of the acid is 300 g/mol and K_f(H₂O) = 1.86^o C/m.

• A. -0.189^oC
• B. -0.194^oC
• C. -0.199^oC
• D. None of these

Answer 1

Answer: (C)

Mass of 1 litre of solution = 1010 g

Mass of solvent $=1010-300\times \text{0.1}\Rightarrow 980g$

$\text{Molality(m)}=\frac{\text{0.1}}{\text{0.98}}\Rightarrow \text{0.102};$

$\Delta {\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\cdot \mathrm{m}\cdot \mathrm{i}=(1+\alpha ){\mathrm{K}}_{\mathrm{f}}\cdot \mathrm{m}$

$\text{Δ}{\text{T}}_{\text{f}}=\text{1.05}\times \text{1.86}\times \text{0.102}={\text{0.199}}^{\text{o}}C;$

${\text{T}}_{\text{f}}=0-\text{0.199}=-{\text{0.199}}^{\text{o}}\text{C}$