The vapour pressure of pure benzene at a certain temperature is 0.500 bar. A non-volatile, non-electrolyte solid weighing 0.5 g is 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 ?

The vapour pressure of pure benzene at a certain temperature is 0.500

1.	The vapour pressure of pure benzene at a certain temperature is 0.500 bar. A non-volatile, non-electrolyte solid weighing 0.5 g is 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» SOLUTION :Here, we are GIVEN that `w_(2)=0.5 g, w_(1)=39.0g, M_(1)="78 g mol"^(-1)` `p^(@)="0.850 BAR, "p_(s)="0.845 bar"` SUBSTITUTING these VALUES in the formula, `(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))=(w_(2)M_(1))/(w_(1)M_(2))`, we get `=("0.850 bar"-"0.845 bar")/("0.850 bar")=(0.5 gxx"78 g mol"^(-1))/(39.0g xxM_(2))` `"or"M_(2)=(0.5xx78)/(39.0)xx(0.850)/(0.005)" g mol"^(-1)="170 g mol"^(-1)`