Question

To a one litre solution of $0.1N$ $HCl$, $0025mol$ of ${NH}_{4}Cl$ are added. Assuming $80$% dissociation of the solutes, the freezing point of the solution is: $(K_f=1.85deg/molal)$

• A. $-{0.33}^{o}C$
• B. $-{0.85}^{o}C$
• C. $-{0.23}^{o}C$
• D. $-{0.416}^{o}C$

Answer 1

The correct option is C $-{0.23}^{o}C$

$\begin{array}{c}\text{0.1 N HCl,}0.025mol{NH}_{4}Cl\\ \Rightarrow 80\%\text{dissociation.}\\ \Rightarrow \alpha =0.8\\ \Rightarrow \frac{i-1}{n-1}=0.8\\ \Rightarrow i=\left(0.8\right)(2-1)+1\\ \Rightarrow i=1.8\end{array}$

$\begin{array}{cc}\Delta T_f& =K_{f}m\\ & =1.86(i\times m_{HCl}+i{m}_{N{H}_{4}Cl})\\ & =1.86(0.1+1.8\times 0.025)\\ & =1.86\left(0.145\right)\\ & =0.2697\end{array}$