Question

The latent heat of fusion of ice is 80 calories per gram at $0^o$C. What is the freezing point of a solution of KCl in water containing 7.45 grams of solute 500 grams of water, assuming that the salt is dissociated to the extent of 95%?

• A. $T_f=-{0.73}^{\circ }C$
• B. $T_f=-{0.92}^{\circ }C$
• C. $T_f=-{1.46}^{\circ }C$
• D. $T_f=-{1.84}^{\circ }C$

Answer 1

The correct option is A $T_f=-{0.73}^{\circ }C$

We know,

$K_f\left(H_{2}O\right)=\frac{R{T_f}^2{M}_{\text{solvent}}}{\Delta H_{\text{fusion}}m\times 1000}$

$=\frac{2\times (273.15{)}^2\times 18}{80\times 18\times 1000}$

$=\frac{2\times 74610.9}{80\times 1000}$

$=1.865K.Kg/mol$

$T_f^{\prime }=0^{o}C-i{K}_f.m$

$=0-\left(1.95\right)\left(1.865\right)\left(\frac{7.45/74.5}{0.5}\right)$

$=-{0.73}^{o}C$

Hence, the correct option is $\text{A}$