The freezing point of a 1.00 , m aqueous solu | vedclass

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Question

(A) Given: Molality $(m) = 1.00 \, m$,$\Delta T_f = 0 - (-1.91) = 1.91 \, K$,$K_f = 1.86 \, K \, kg \, mol^{-1}$.Using the formula: $\Delta T_f = i

Vedclass · Accepted Answer

$2.7$

Vedclass · Answer

(A) Given: Molality $(m) = 1.00 \, m$,$\Delta T_f = 0 - (-1.91) = 1.91 \, K$,$K_f = 1.86 \, K \, kg \, mol^{-1}$.Using the formula: $\Delta T_f = i 	imes K_f 	imes m$.$1.91 = i 	imes 1.86 	imes 1.00 \Rightarrow i = \frac{1.91}{1.86} \approx 1.02688$.For dissociation $HF ightleftharpoons H^+ + F^-$,the number of ions $(n) = 2$.The degree of dissociation $(\alpha)$ is given by $\alpha = \frac{i - 1}{n - 1}$.$\alpha = \frac{1.02688 - 1}{2 - 1} = 0.02688$.Percentage of dissociation $= 0.02688 	imes 100 \approx 2.7 \%$.

The freezing point of a $1.00 \, m$ aqueous solution of $HF$ is $-1.91^o C$. The freezing point depression constant of water $K_f$ is $1.86 \, K \, kg \, mol^{-1}$. The percentage of dissociation of $HF$ at this concentration is ......... $\%$

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