The freezing point of a 0.262 mol kg^{-1} s | vedclass

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(B) Given:Molality $(m)$ = $0.262 \ mol \ kg^{-1}$Freezing point of solution $(T_f)$ = $277.4 \ K$Freezing point of pure benzene $(T_f^\circ)$ = $278.

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$0.76$

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(B) Given:Molality $(m)$ = $0.262 \ mol \ kg^{-1}$Freezing point of solution $(T_f)$ = $277.4 \ K$Freezing point of pure benzene $(T_f^\circ)$ = $278.4 \ K$$K_f$ for benzene = $5.0 \ K \ kg \ mol^{-1}$Step $1$: Calculate the depression in freezing point $(\Delta T_f)$:$\Delta T_f = T_f^\circ - T_f = 278.4 \ K - 277.4 \ K = 1.0 \ K$Step $2$: Calculate the theoretical depression in freezing point $(\Delta T_f(	ext{theoretical}))$:$\Delta T_f(	ext{theoretical}) = K_f 	imes m = 5.0 \ K \ kg \ mol^{-1} 	imes 0.262 \ mol \ kg^{-1} = 1.31 \ K$Step $3$: Calculate the van't Hoff factor $(i)$:$i = \frac{\Delta T_f(	ext{observed})}{\Delta T_f(	ext{theoretical})} = \frac{1.0}{1.31} \approx 0.763$Thus,the van't Hoff factor is approximately $0.76$.

The freezing point of a $0.262 \ mol \ kg^{-1}$ solution of acetic acid in benzene is $277.4 \ K$. If the $K_f$ value for benzene is $5.0 \ K \ kg \ mol^{-1}$ and the freezing point of pure benzene is $278.4 \ K$,then the van't Hoff factor is ...........

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