JEE Advance - Chemistry (1990 - No. 11)
The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular weight of the solid substance?
69.6
65.25
72.5
78
60
Explanation
Given, vapour pressure of pure benzene, p0 = 640 mm Hg, vapour pressure of the solution, p = 600 mm Hg, mass of benzene, W1 = 39.0 g, mass of the solid, W2 = 2.175 g
We know, M1 = molar mass of benzene = 78.
According to Raoults' law, $${{{p^0} - p} \over {{p^0}}} = {{{W_2} \times {M_1}} \over {{M_2} \times {W_1}}}$$
$$\therefore$$ $${{640 - 600} \over {640}} = {{2.175 \times 78} \over {{M_2} \times 39}}$$ or, $${M_2} = {{2.175 \times 78 \times 640} \over {39 \times 40}}$$
or, $${M_2} = 69.6$$, i.e., molecular mass of the solid = 69.6.
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