0.440 g of a substance dissolved in 22.2 g | vedclass

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Question

(A) The formula for molar mass $(M_2)$ using depression in freezing point is: $M_2 = \frac{K_f 	imes w_2 	imes 1000}{\Delta T_f 	imes w_1}$ Given:

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$178.9$

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(A) The formula for molar mass $(M_2)$ using depression in freezing point is: $M_2 = \frac{K_f 	imes w_2 	imes 1000}{\Delta T_f 	imes w_1}$ Given: $K_f = 5.12 \ ^oC \ kg \ mol^{-1}$ $w_2 = 0.440 \ g$ $w_1 = 22.2 \ g$ $\Delta T_f = 0.567 \ ^oC$ Substituting the values: $M_2 = \frac{5.12 	imes 0.440 	imes 1000}{0.567 	imes 22.2} = \frac{2252.8}{12.5874} \approx 178.9 \ g \ mol^{-1}$ Thus,the correct option is $A$.

$0.440 \ g$ of a substance dissolved in $22.2 \ g$ of benzene lowered the freezing point of benzene by $0.567 \ ^oC$. Calculate the molecular mass of the substance. (Given: $K_f = 5.12 \ ^oC \ kg \ mol^{-1}$)

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