The vapour pressure of an aqueous solution of glucose is 750 mm mercury at 373 kelvin calculate molality and mole fraction
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At 373K V.P of water is 750 mm Hg
We know that, (P0-P) / P0 = w x M / m x W
molality = (P0-P) / P0 X 1000/M
=> (760-750) / 760 X 1000/18
=> 0.73 m
Answer>> Mole fraction = (P0-P) / P0 = 10/760 = 0.0131
We know that, (P0-P) / P0 = w x M / m x W
molality = (P0-P) / P0 X 1000/M
=> (760-750) / 760 X 1000/18
=> 0.73 m
Answer>> Mole fraction = (P0-P) / P0 = 10/760 = 0.0131
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