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An aqueous solution of 2% non-volatile s...

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent . What is the molecular mass of the solute ?

Text Solution

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Vapour pressure of pure solvent at its boiling point, `p_(A)^(0)= 1` atm =1.013 bar
Vapour pressure of solution, p = 1.004 bar
Weight of solute, `W_(1)=2g, M_(2)=`?
Weight of solvent `= (100-2) = 98` g
`M_(A)=18" g mol"^(-1)` (Mol. mass of water)
Now `(p^(0)-p)/(p_(A)^(0))=(W_(2) xx M_(1))/(W_(1) xx M_(2))`
or `(1.013-1.004)/(1.013)=(2)/(98)xx(18)/(M_(2))`
or `(0.009)/(1.013) = (2xx18)/(98xxM_(2))`
`M_(2)=(2xx18xx1.013)/(0.009 xx 98)=41.3" g mol"^(-1)`
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