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The vapour pressure of pure benzene at a...

The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5g when added to 39.0 g of benzene (molar mass 78 g `"mol"^(-1)`), vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance ?

A

180

B

270

C

160

D

169

Text Solution

Verified by Experts

The correct Answer is:
D
If, `p^ (@)= ` vapour pressure of pure benzene . P= vapour pressure of solution.
` (p ^(@) -p)/(p) = (n(solute ))/(n(solvent))= (n_(1))/(n_(2))=(0.850.845)/(0.845)=(W_(1))/(M_(1))xx(M_(2))/(W_(2))`
where , `w_(1)` and `w_(2)` are masses of solute respectively and `M_(1),M_(2)` are molar masses of solute and solvent respectively.
`M_(1) = (0.5xx78xx0.845)/(0.05xx39)`
`M _(1) = 169g mol^(-)`
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