MHT CET · Chemistry · Solutions

The vapour pressure of benzene at a certain temperature is \(640 \mathrm{~mm}\) of \(\mathrm{Hg} .\) A non-volatile and non-electrolyte solid weighing \(2.175 \mathrm{~g}\) is added to \(39.08 \mathrm{~g}\) of benzene. If the vapour pressure of the solution is \(600 \mathrm{~mm}\) of \(\mathrm{Hg}\), what is the molecular weight of solid substance?

A \(49.50\)
B \(59.60\)
C \(69.60\)
D \(79.82\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(69.60\)

Step-by-step Solution

Detailed explanation

Given, vapour pressure of benzene,
\(
p^{\circ}=640 \mathrm{~mm} \mathrm{Hg}
\)
Vapour pressure of solution,
\(
p=600 \mathrm{~mm} \mathrm{Hg}
\)
Weight of solute, \(w=2.175 \mathrm{~g}\) Weight of benzene, \(W=39.08 \mathrm{~g}\) Molecular weight of benzene,
\(
M=78 \mathrm{~g}
\)
Molecular weight of solute, \(m=?\) According to Raoult's law,
\(\frac{p^{\circ}-p}{p^{\circ}} =\frac{w \times M}{m \times W} \)
\( \frac{640-600}{640} =\frac{2.175 \times 78}{m \times 39.08} \)
\( \frac{40}{640} =\frac{2.175 \times 78}{m \times 39.08} \)
\( m =\frac{16 \times 2.175 \times 78}{39.08} \)
\( m =69.60\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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