Question

The vapour pressure of pure benzene is 640mm of hg.2.175 x 10-3 kg of non-volatile solute is added to 39gram of benzene. The vapour pressure of solution is 600mm of hg. Calculate molar mass of solute. (c=12,h=1)

Answer 1

Hello Student,

Please find the answer to your question

According to Rault’s law,

Po-­P/Po = w/m /w/m +W/M

Here, Po = 640 mm p = 600 mm

W = 2.175 g W = 39.0

M = ? M = 78

Substituting the various values in the above equation for Roult’slaw :

640 -600/640 = 2.175/m /2.175/m+39/78

1/16 = 2.175/2.175 + 0.5 m

m = 65.25