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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

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Accoridng to Raoult's Law ,`(P_(A)^(@)-P_(S))/P_(S)=n_(B)/n_(A)=W_(B)/M_(B)xxM_(A)/W_(A)`
`P_(A)^(@)=0.850 bar,P_(S)=0.845 bar, W_(B)=0.5 g W_(A)=39.0g`
`M_(A)78g mol(-1)`
`((0.850-0.845))/((0.845 bar))=((0.5g)xx(78 gmol^(-1)))/(M_(B)xx(39.0g))`
`M_(B)=((0.5g)xx(78 gmol^(-1)))/((39.0 g)xx(0.005 bar))=169 g mol^(-1)`
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