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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 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?

Text Solution

Verified by Experts

Accroding to Roult.s law .
`(p_(A)^(@) - p_(A))/(p_(A)) = x_(B) (w_(B) xx M_(A))/(M_(B) xx W_(B))` ( for dilute solution )
`w_(B) = 0.50 g, w_(A) = 39.0 g,p_(A) = 0.845 ` bar `p_(A)^(@) = 0.580` bar
`(0.850 - 0.845)/(0.850) = (0.5 xx 78 )/(39.0 xx M_(B))`
`(0.005)/(0.850) = (0.5 xx 78)/(39.0 xx M_(B))`
or ` M_(B) = (0.5 xx 78 xx 0.850 )/(0.005 xx 39.0) = 170 g mol^(-1)`
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