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The vapour pressure of a 10% aqueous sol...

The vapour pressure of a 10% aqueous solution of a non volatile substance at 373K is 740 mm of Hg. Calculate the molecular mass of the solute.

Text Solution

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Mass of solute= 10g. Mass of water= 90g
Vapour pressure of solution at 373K= 740mm
Moles of water in 90g water= `(90)/(18)`= 5 moles
Relative lowering of vapour pressure `(P_(A)^(0)- P_(s))/(P_(A)^(0))= (n_(B))/(n_(A) + n_(B)) (760-740)/(760)= (n_(B))/(5+n_(B)) rArr (20)/(760) = (n_(B))/(5 +n_(B))`
`20 xx 5` mol `+20n_(B)= 760n_(B) rArr 760n_(B)-20 n_(B)= 20 xx 5` mol
`n_(B)= (20 xx 5)/(740)= 0.135` mol
`n_(B)= ("Mass of solute")/("Molar mass of solute")rArr 0.135 mol= (10g)/(M_(B))`
Molar mass of solute, `M_(B)= (10g)/(0.135mol)= 74.07 g mol^(-1)`
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