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The vapour pressure of pure benzene at 5...

The vapour pressure of pure benzene at `50^(@)` is `268 mm` of `Hg`. How many moles of non-volatile solute per mole of benzene are required to prepare a solution of benzene having a vapour pressure of `16.0 mm` of `Hg` at `50^(@)C`?

Text Solution

Verified by Experts

Applying Raoult's law in the following form:
`(P^(@)-P_(S))/(P_(S))=(W_(2)//Mw_(2))/(W_(1)//Mw_(1))`
= Number of moles of solute per mole of benzene
or `n_(2) / n_(1) = (286 - 160) /160 = 0.675`
Alternative method
We know that `P_(S)` = Mole of solvent `xx P^(@)`
or `160` = Mole fraction of solvent `xx 268`
So, molar fraction of solvent = `160/268 = 0.597`
Mole fraction of solute =`1-0.597 =0.403`
`n_(2) /n_(1) = ("Mole fraction of solute" )/("Mole fraction of solvent") = 0.403 / 0.597 = 0.675`
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