Q. The vapour pressure of benzene at a certain temperature is 640 mm of Hg. A non-volatile and non-electrolyte solid weighing 2.175 g is added to 39.08 g of benzene. If the vapour pressure of the solution is 600 mm of Hg, what is the molecular weight of solid substance?
A) 49.50
B) 59.60
C) 69.60
D) 79.82
Explanation
(c)
Given, vapour pressure of benzene,
$p^{\circ}$=640 mm Hg
Vapour pressure of solution,
p=600 mm Hg
Weight of solute, w= 2.175 g
Weight of benzene, W= 39.08 g
Molecular weight of benzene,
M=78 g
Molecular weight of solute, m=?
According to Raoult’s law,
$\frac{P^{\circ} - P}{P^{\circ}} = \frac{w \times M}{m \times W}$
$\frac{640 - 600}{640} = \frac{2.175 \times 78}{m \times 39.08}\ \ \$
$\frac{40}{640} = \frac{2.175 \times 78}{m \times 39.08}$
m=$\frac{16 \times 2.175 \times 78}{39.08}$
m=69.60
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.