MHT CET · Chemistry · Solutions
Find the molar mass of solute when 2 gram dissolved in 60 gram benzene at \(30^{\circ} \mathrm{C}\) and relative lowering of vapour pressure is 0.06 . (Molar mass of benzene is \(78 \mathrm{~g} \mathrm{~mol}\) )
- A 17.4 gram \(\mathrm{mol}^{-1}\)
- B 35.2 gram mol-1
- C 43.3 gram \(\mathrm{mol}^{-1}\)
- D 24.2 gram mol-1
Answer & Solution
Correct Answer
(C) 43.3 gram \(\mathrm{mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\frac{\Delta \mathrm{P}}{\mathrm{P}_{\mathrm{A}}^{\circ}}=\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}} \times \frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{A}}}\)
\(0.06=\frac{2}{M_B} \times \frac{78}{60}\)
\(M_B=\frac{78}{30 \times 0.06}=43.3 \mathrm{~g} / \mathrm{mol}\)
\(0.06=\frac{2}{M_B} \times \frac{78}{60}\)
\(M_B=\frac{78}{30 \times 0.06}=43.3 \mathrm{~g} / \mathrm{mol}\)
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