Question

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

Answer 1

the various quantities known to us are as follows
`p_(1)^(0)=0.850` bar: `p=0.845` bar, `M_(1)=78g" "mol^(_1),w_(2)=0.5g,w_(1)=39g`
substituting these values in equation , we get
`(0.850" bar"-0845" bar")/(0.850" bar")=(0.5gxx78" g "mol^(-1))/(M_(2)xx39g)`
therefore, `M_(2)=170g" "mol^(-1)`