Question

A $0.1M$ aqueous solution of a monoprotic acid $(\rho =1.01g/c{m}^3)$ is $5\%$ ionized. Calculate the freezing point of the solution.
Given:
$\text{Molecular weight of acid}=300gmo{l}^{-1}$

$K_f\left(H_{2}O\right)={1.86}^{\circ }C/m:$

• A. $-{0.399}^{\circ }C$
• B. $+{0.199}^{\circ }C$
• C. $-{0.199}^{\circ }C$
• D. ${0.399}^{\circ }C$

Answer 1

The correct option is C $-{0.199}^{\circ }C$

Since , $\rho =1.01g/mL$
$\text{Mass of 1 litre of solution}=1010g$

$0.1M$ means 0.1 moles of solute are there in one litre of solution.
$\text{Mass of solute}=\text{Molecualr mass of acid}\times 0.1=300\times 0.1=30g$
$\text{Mass of solvent}=(1010-30)g=980g$

$\text{Molality}\left(m\right)=\frac{0.1\times 1000}{980}=0.102$
For monoprotic acid, $HA$
$n=2\text{Vant Hoff factor}\left(i\right)=1-\alpha +n\alpha i=1+\alpha i=1.05$

$\Delta T_f=i\times K_f\times m$
$\Delta T_f=1.05\times 1.86\times 0.102={0.199}^{\circ }C$
$T_f^0-T_f=0.199$
$T_f^0\text{is the freezing pt. of pure water}{T}_f\text{is the freezing pt of solution}$
$T_f=(0-0.199{)}^{\circ }C=-{0.199}^{\circ }C$