Question

An aqueous solution contains 5% by mass of urea and 10% by mass of glucose. If ${\text{K}}_{\text{f}}$ for water is $\text{1.86}{\text{K kg}}^{}{\text{mol}}^{-1},$ the freezing point of the solution is

• A. $-3.03\text{K}$
• B. $3.03\text{K}$
• C. $-3.03{}^{\text{o}}\text{C}$
• D. $3.03{}^{\text{o}}\text{C}$

Answer 1

Answer: (C)

Moles of solute $=\frac{5}{60}+\frac{10}{180}=\frac{25}{180}$
$\text{Molality}\text{=}\frac{\text{n}\text{×}\text{1000}}{\text{weight}\text{of}\text{solvent}}$
Molality of solution $=\frac{25\times 1000}{180\times 85}$
$\mathrm{\Delta }{\mathrm{T}}_{\mathrm{f}}=\frac{25\times 1000\times 1.86}{180\times 85}=3.03$
$F.P.=0-3.03{}^{\text{o}}\text{C}=-3.03{}^{\text{o}}\text{C}$