the vapour pressure of pure benzene at a certain temperature is 640mmhg a non volatile solute of mass 2.175*10-3 kg is added to 39.0*10-3 kg of benzene . the vapour pressure of the solution is 600mmhg.whatis the molar mass of the solute? 
 

Asked by maitridesai429 | 2nd Mar, 2015, 01:05: PM

Expert Answer:

By the Raoult's Law,
 
x1 = P1 / P1° = 600 mmHg / 640 mmHg = 0.9375
 
Also, x1 = n1 / n1 + n2 = (39/78) / (39/78) + (2.175/M)
 
0.9375 = (39/78) / (39/78) + (2.175/M) = 65.91 g/mol

Answered by Arvind Diwale | 3rd Mar, 2015, 11:17: AM