Question

To $500$ $c{m}^3$ of water, $3.0\times {10}^{-3}$ kg of acetic acid is added. If $23\%$ of acetic acid is dissociated, what will be the depression in freezing point. If $k_f$ and density of water are $1.86$ $K$ $k{g}^{-1}$ and $0.997$ $gc{m}^{-3}$ respectively.

• A. $0.186$ K
• B. $0.228$ K
• C. $0.372$ K
• D. $0.556$ K

Answer 1

The correct option is B $0.228$ K

Number of moles of acetic acid= $\frac{3\times {10}^{-3}kg}{60gmo{l}^{-1}}=\frac{3g}{60gmo{l}^{-1}}=0.05$

Mass of water= $m_1=500c{m}^3\times 0.997gc{m}^{-3}=0.4985kg$

Molality of acetic acid= $m=\frac{n_2}{m_1}=\frac{0.05mol}{0.4985kg}=0.1003molk{g}^{-1}$

Dissociation of $C{H}_{3}COOH$:

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{⊝}+H^{\oplus }$

$t=0$ $1$ $0$ $0$

eqm $1-\alpha$ $\alpha$ $\alpha$

Total number of moles= $i=1-\alpha +\alpha +\alpha =1+\alpha$

as, $\Delta T_f=i{K}_{f}m$

$\Delta T=(1+\alpha )K_{f}m=(1+0.23)\left(1.86KKgmo{l}^{-1}\right)\left(0.1003molK{g}^{-1}\right)=0.228K$

Hence, the correct option is $B$