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The vapour pressure of a dilute aqueous ...

The vapour pressure of a dilute aqueous solution of glucose is `700 mm` of `Hg` at `373 K`. Calculate the (a) molality and (b) mole fraction of the solute.

Text Solution

Verified by Experts

`P^(@)` = Vapour pressure of water at `373 K` = `760 mm Hg`
Using Raoult's law in the following form,
`(P^(@) - P^(S))/ P^(S) = (W_(2)Mw_(1))/ (W_(1)Mw_(2))`
or `(760-700)/(700) = (W_(2) xx 18)/(W_(1) xx Mw_(2))`
or `(W_(2))/(W_(1)xxMw_(2))=(60)/(700xx18)`
Molality = `(W_(2))/(W_(1) xxMw_(2)) xx 1000 = (60 xx 1000)/(700 xx 18) = 4.76 m`
`P_(S) =` Mole fraction of solvent `xx P^(@)`,
Mole fraction of solvent `= (700)/(760) =0.921`
So, mole fraction of solute `= 1-0.921 = 0.0789`
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