The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78g `mol^(-1)`). Vapour pressure of the solution, then is 0.845 bar. What is the molar mass of the solid substance?

Answer

170 g mol

Explanation

`p^(0)`=0.850 bar, p = 0.845 bar = 0.5 g , 845 bar
Weight of solute = 0.5 g , Weight of benzène = 39.0 g
Molecular mass of benzene `= 78" g mol"^(-1)`
No. of moles of solute `(n_(2))=(0.5)/(M)`
No. of moles of benzene `(n_(1))=(39)/(78)=0.5`
Now, `(p^(0)-p)/(p^(0))=(n_(2))/(n_(1)) or (0.850-0.845)/(0.850)=(0.5)/(M xx 0.5)=(1)/(M)`
or `(0.005)/(0.850)=(1)/(M) or M=(0.850)/(0.005)=170"g mol"^(-1)`