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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?

A

170 amu

B

160 amu

C

120 amu

D

190 amu

Text Solution

Verified by Experts

The correct Answer is:
1
we are given
`P_(1)^(0)=0.850` bar, `P_(1)`=0.845 bar
`M_(1)=78 g mol^(-1), W_(1)=39 g and W_(2)=0.5 g`
Substituting these values in Equation `(2.58)`, we get
`DeltaP_(1)//P_(1)^(0)=(W_(2)xxM_(1))/(W_(1)xxM_(2))`
`((0.850 - 0.845) "bar")(0.850 "bar")=(0.5 gxx78 g mol^(-1))/(M_(2)xx39 g)`
or `M_(2)=170 g mol^(-1)`
Thus, molecular mass of noncolatile solid is 170 amu
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