Question

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?

Answer 1

The given values are
`P^(@) = 0.850 bar` , `P_(S) = 0.845 bar` , `Mw_(1) = 78 g mol^(-1)`
`W_(2) = 0.5 g`, `W_(1) = 39 g`
Now using equation
`= (P^(@) - P_(S)) / P^(@) = (W_(2) xx Mw_(1)) / (Mw_(2) xx W_(1))`
On substituting all the given values, we get
`(0.850 - 0.845) /0.850 = (0.5 g xx 78 g mol^(-1)) /(Mw_(2) xx 39 g)`
`:. Mw_(2) = 170 g mol_(-1)`