1.95 g of non-volatile and non-electrolyte s | vedclass

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(B) The formula for freezing point depression is $\Delta T_{f} = K_{f} 	imes m$,where $m$ is the molality of the solution. Molality $m = \frac{w_{sol

Vedclass · Accepted Answer

$156$

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(B) The formula for freezing point depression is $\Delta T_{f} = K_{f} 	imes m$,where $m$ is the molality of the solution. Molality $m = \frac{w_{solute} 	imes 1000}{M_{solute} 	imes w_{solvent(g)}}$. Given: $\Delta T_{f} = 0.64 \ K$,$w_{solute} = 1.95 \ g$,$w_{solvent} = 100 \ g$,$K_{f} = 5.12 \ K \ kg \ mol^{-1}$. Substituting the values: $0.64 = 5.12 	imes \frac{1.95 	imes 1000}{M_{solute} 	imes 100}$. $0.64 = 5.12 	imes \frac{19.5}{M_{solute}}$. $M_{solute} = \frac{5.12 	imes 19.5}{0.64}$. $M_{solute} = 8 	imes 19.5 = 156 \ g \ mol^{-1}$.

$1.95 \ g$ of non-volatile and non-electrolyte solute dissolved in $100 \ g$ of benzene lowered the freezing point of it by $0.64 \ K$. The molar mass of the solute (in $g \ mol^{-1}$) is: $(K_{f}(C_6H_6) = 5.12 \ K \ kg \ mol^{-1})$

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