Question

Vapour pressure of benzene at some temperature is 0.95 bar. Calculate the vapour pressure of 1 molal solution of a non volatile solute (i=1) in it. Given molar mass of benzene is 78g/mol

Answer 1

Answer:

Pi° =0.850 bar:

p=0.845 bar:

M1 = 78 g mol-1

w2 = 0.5 g

w1=39g

Substituting these values In equation .

and calculating,

M2= 170 g mol-1 3 years ago

P°=0.850

P'=0.845

No of moles of benzene=39/78=0.5

No of moles of solute=x

(P°-P')/p°=mole fraction of solute

P°-p'=0.005

Mole fraction of solute=0.005/0.850----*1

Mole fraction of solute=x/(x+0.5)-----*2

Sub eq 1 and 2

X=169g

Answer 2

vapour pressure of pure benzene at some temperature is 0.95bar calculate the vapour pressure of 1 molar solution of a non-volatile solute in it (given molar mass of benzene - 78g/mol )