JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
\(1.22\, \mathrm{~g}\) of an organic acid is separately dissolved in \(100 \,\mathrm{~g}\) of benzene \(\left(\mathrm{K}_{\mathrm{b}}=2.6\, \mathrm{~K}\, \mathrm{~kg} \,\mathrm{~mol}^{-1}\right)\) and \(100\, \mathrm{~g}\) of acetone \(\left(\mathrm{K}_{b}=1.7\, \mathrm{~K} \,\mathrm{~kg} \,\mathrm{~mol}^{-1}\right) .\) The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by \(0.17^{\circ} \mathrm{C}\). The increase in boiling point of solution in benzene in \({ }^{\circ} \mathrm{C}\) is \(\mathrm{x} \times 10^{-2}\). The value of \(\mathrm{x}\) is ..... .(Nearest integer) \([\) Atomic mass : \(\mathrm{C}=12.0, \mathrm{H}=1.0, \mathrm{O}=16.0]\)
- A \(12\)
- B \(13\)
- C \(10\)
- D \(11\)
Answer & Solution
Correct Answer
(B) \(13\)
Step-by-step Solution
Detailed explanation
With benzene as solvent \(\Delta \mathrm{T}_{b}=\mathrm{i} \mathrm{K}_{\mathrm{b}} \mathrm{m}\) \(\Delta \mathrm{T}_{\mathrm{b}}=\frac{1}{2} \times 2.6 \times \frac{1.22 / \mathrm{M}_{\mathrm{w}}}{100 / 1000}...(1)\) With Acetone as solvent…
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