A solution of 5 g nonvolatile solute in 50 | vedclass

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

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$930 \ g \ mol^{-1}$

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(B) The formula for depression in freezing point 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: $\Delta T_{f} = 0.2 \ K$,$W_{solute} = 5 \ g$,$W_{solvent} = 50 \ g$,$K_{f} = 1.86 \ K \ kg \ mol^{-1}$.Substituting the values: $0.2 = 1.86 	imes \frac{5 	imes 1000}{M_{solute} 	imes 50}$.$0.2 = 1.86 	imes \frac{100}{M_{solute}}$.$M_{solute} = \frac{1.86 	imes 100}{0.2} = \frac{186}{0.2} = 930 \ g \ mol^{-1}$.

$A$ solution of $5 \ g$ nonvolatile solute in $50 \ g$ water decreases its freezing point by $0.2 \ K$. Calculate the molar mass of solute if $K_{f}$ of water is $1.86 \ K \ kg \ mol^{-1}$.

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