$$100$$ g of $$C6H12O6(aq)$$ solution has vapour pressure is equal to $$40$$ torr at certain temperature. Vapour pressure of $$H2O(1)$$ is $$40.18$$ torr at same temperature. If this solution is cooled to $$-0.93$$°C, what mass of ice will be separeted out? (Kf$$ $$=$$ $$1.86$$ kg $$mol-1)

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Infinity Learn · Accepted Answer

Molality of solution $$=$$ P°$$-PP$$×$$1000M$$ ⇒$$40.18-4040$$×$$100018$$⇒$$0.251000$$ g water present with $$45$$ g glucose or $$100$$ g solution has $$4.31$$ g glucose and $$95.69$$ g $$H2O$$. Final molality is $$0.5$$, $$1000$$ g solvent contain $$90$$ g glucose or $$4.31$$ g glucose present with $$100090$$×$$4.31$$ $$=$$ $$47.88$$ $$gH2O$$ Mass of ice formed $$=$$ $$95$$ $$.69$$ $$-$$ $$47$$ $$.88$$ $$=$$ $$47$$ $$.8$$ g

100 g of C₆H₁₂O₆(aq) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H₂O(1) is 40.18 torr at same temperature. If this solution is cooled to -0.93 $°$ C, what mass of ice will be separeted out? (K_{f =}1.86 kg mol^-1)

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100 g of C6H12O6(aq) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H2O(1) is 40.18 torr at same temperature. If this solution is cooled to -0.93°C, what mass of ice will be separeted out? (Kf = 1.86 kg mol-1)

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Ready to Test Your Skills?

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100 g of C₆H₁₂O₆(aq) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of H₂O(1) is 40.18 torr at same temperature. If this solution is cooled to -0.93 $°$ C, what mass of ice will be separeted out? (K_{f =}1.86 kg mol^-1)