The vapour pressure of an aqueous solution of glucose is 750 mm of mercury at `100^(@)C`. Calculate the molatlity and mole fraction of solute.

The vapour pressure of an aqueous solution of glucose is 750 mm of mer

1.	The vapour pressure of an aqueous solution of glucose is 750 mm of mercury at `100^(@)C`. Calculate the molatlity and mole fraction of solute.
Answer» Given that temperature is `372 K` and `b`. Pt. of `H_(2)O = 373 K` `:.` Vapour pressure of `H_(2)O = 76 cm` We have, `(P^(@)-P_(S))/(P_(S)) = ( w xx M)/(mxxW)` `:.` Molality `= (w)/(mxx W) xx 1000 = (P^(@)-P_(S))/(P_(S)) xx (1)/(M) xx 1000` `= (760-750)/(750) xx (1)/(18) xx 1000` `= 0.741 mol//kg` of solvent Also we have, `(P^(@)-P_(S))/(P^(@)) = (n)/(n+N)` `:.` Mole fraction `= (P^(@)-P_(S))/(P^(@)) = (760-750)/(760) = (10)/(760) = 0.013`

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The vapour pressure of an aqueous solution of glucose is 750 mm of mercury at `100^(@)C`. Calculate the molatlity and mole fraction of solute.

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