Question

The vapour pressure of pure benzene at a certain temperature is 0.850 bar.

Answer 1

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

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