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Question: Answered & Verified by Expert
There are following the three solvents \( \mathrm{X}, \mathrm{Y} \& \mathrm{Z} \). They have identical molar masses. Match their boiling point with their \( \mathrm{k}_{\mathrm{b}} \)
values:
\( \begin{array}{ccc}\text { Solvent } & \text { B. Pt. } & \mathrm{K}_{\mathrm{b}} \\ \mathrm{X} & 100^{\circ} \mathrm{C} & 0.92 \\ \mathrm{Y} & 27^{\circ} \mathrm{C} & 0.63 \\ \mathrm{Z} & 283^{\circ} \mathrm{C} & 0.53\end{array} \)
Among the following, which is the correct relation between the boiling point and \( \mathrm{K}_{\mathrm{b}} \) values for \( \mathrm{X}, \mathrm{Y}, \mathrm{Z} \) respectively?
ChemistrySolutionsJEE Main
Options:
  • A \( 100^{\circ} \mathrm{C}-0.92,27^{\circ} \mathrm{C}-0.63,283^{\circ} \mathrm{C}-0.53 \)
  • B \( 100^{\circ} \mathrm{C}-0.53,27^{\circ} \mathrm{C}-0.63,283^{\circ} \mathrm{C}-0.92 \)
  • C \( 100^{\circ} \mathrm{C}-0.63,27^{\circ} \mathrm{C}-0.53,283^{\circ} \mathrm{C}-0.92 \)
  • D \( 100^{\circ} \mathrm{C}-0.53,27^{\circ} \mathrm{C}-0.92,283^{\circ} \mathrm{C}-0.63 \)
Solution:
1362 Upvotes Verified Answer
The correct answer is: \( 100^{\circ} \mathrm{C}-0.63,27^{\circ} \mathrm{C}-0.53,283^{\circ} \mathrm{C}-0.92 \)
K b = MRT b 2 1 0 0 0 Δ H vap
K b M T b Δ H vap T b
K b T b Δ S vap
Δ S  will be same for three liquids changing to vapours for 1 mole
K b T b

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