5The vapour pessure of pure benzene is 640 mm of Hg. 2.175 x 103 kg of

1.	5The vapour pessure of pure benzene is 640 mm of Hg. 2.175 x 103 kg of nonvolatilesolute is added to 39 gram of benzene the vapour pressure of solution is 600 mm of Hg.Calculate molar mass of solute.
Answer» 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 Like my answer if you find it useful!

5The vapour pessure of pure benzene is 640 mm of Hg. 2.175 x 103 kg of nonvolatilesolute is added to 39 gram of benzene the vapour pressure of solution is 600 mm of Hg.Calculate molar mass of solute.

Answer»

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

Like my answer if you find it useful!