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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 ?

Text Solution

Verified by Experts

The various quantities known to us are as follows.
`P_(1)^(0)=0.850" bar"`
`P=0.845"bar"`
`M_(1)=78" g mol"^(-1)`
`w_(2)=0.5" g"`
`w_(1)=39"g"`
Substituting these values in equation `(P^(0)-P)/(P_(1)^(0))=(w_(2)xxM_(1))/(M_(2)xx w_(1))` we get
`(0.850" bar"-0.845" bar")/(0.850" bar")=(0.5"g"xx78" g mol"^(-1))/(M_(2)xx39" g")`
Therefore, `M_(2)=170" g mol"^(-1)`.
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