1 kg of 0.75 m aqueous solution of sucrose | vedclass

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(C) Initial molality $m = 0.75 \ mol \ kg^{-1}$.Let the mass of water be $w_{1} \ kg$ and mass of sucrose be $w_{2} \ kg$.Total mass $= w_{1} + w_{2}

Vedclass · Accepted Answer

$518$

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(C) Initial molality $m = 0.75 \ mol \ kg^{-1}$.Let the mass of water be $w_{1} \ kg$ and mass of sucrose be $w_{2} \ kg$.Total mass $= w_{1} + w_{2} = 1 \ kg$.Molality $m = \frac{w_{2} / 342}{w_{1}} = 0.75 \implies w_{2} = 0.75 	imes 342 	imes w_{1} = 256.5 	imes w_{1}$.Substituting into $w_{1} + 256.5 	imes w_{1} = 1 \implies 257.5 	imes w_{1} = 1 \implies w_{1} \approx 0.003883 \ kg$ (This interpretation is incorrect based on the problem statement).Correct approach: $0.75 \ m$ solution means $0.75 \ mol$ sucrose in $1 \ kg$ water. Total mass $= 1000 + (0.75 	imes 342) = 1256.5 \ g$.Mass of sucrose in $1000 \ g$ solution $= \frac{0.75 	imes 342}{1256.5} 	imes 1000 \approx 204.14 \ g$.Mass of water in $1000 \ g$ solution $= 1000 - 204.14 = 795.86 \ g$.Moles of sucrose $= 0.75 	imes 0.79586 = 0.5969 \ mol$.Freezing point depression $\Delta T_{f} = 4 = K_{f} 	imes m_{new} = 1.86 	imes \frac{0.5969}{w_{new(kg)}}$.$w_{new} = \frac{0.5969 	imes 1.86}{4} = 0.2775 \ kg = 277.5 \ g$.Ice separated $= 795.86 - 277.5 = 518.36 \ g \approx 518 \ g$.

$1 \ kg$ of $0.75 \ m$ aqueous solution of sucrose is cooled to $-4^{\circ} C$. The amount of ice (in $g$) that will be separated out is .... . (Nearest integer)
Given: $K_{f}(H_{2}O) = 1.86 \ K \ kg \ mol^{-1}$

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$1 \ kg$ of $0.75 \ m$ aqueous solution of sucrose is cooled to $-4^{\circ} C$. The amount of ice (in $g$) that will be separated out is .... . (Nearest integer)Given: $K_{f}(H_{2}O) = 1.86 \ K \ kg \ mol^{-1}$