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Home/Practice questions/Chemistry
ChemistryPhysical: Solutions

Arrange the following aqueous solutions in increasing order of their…

Arrange the following aqueous solutions in increasing order of their boiling points.

Given: Molar mass (in g mol−1) of H, C, N, O, S and K are 1, 12, 14, 16, 32 and 39, respectively.

A. 5% K2SO4
B. 10 % Urea
C. 5% Urea
D. 10% K2SO4

Choose the correct answer from the options given below:

Answer & explanation+

Correct answer: option 2

The correct answer is Option (2) → C, A, B, D

Elevation of boiling point is a colligative property. It depends on the number of solute particles per kilogram of solvent:

    \( \displaystyle \Delta T_b=i\,K_b\,m \)

Since Kb is the same for all four aqueous solutions, compare the effective molality i × m. Two factors decide it:

  • van’t Hoff factor: urea does not dissociate: 1 particle, i = 1. \( \ce{K2SO4 -> 2K+ + SO4^2-} \): 2 + 1 = 3 ions, i = 3.

  • Moles per kg of water: urea has a much smaller molar mass (60 g mol−1) than \( \ce{K2SO4} \) (2 × 39 + 32 + 4 × 16 = 174 g mol−1), so the same mass gives more moles.

An x% solution by mass has x g of solute in (100 − x) g of water, so the 5% solutions have 0.095 kg and the 10% solutions 0.090 kg of water per 100 g:

    \( \displaystyle \text{C, urea}:\quad i\,m=1\times\frac{5/60}{0.095}=0.877 \)

    \( \displaystyle \text{A, }\ce{K2SO4}:\quad i\,m=3\times\frac{5/174}{0.095}=0.907 \)

    \( \displaystyle \text{B, urea}:\quad i\,m=1\times\frac{10/60}{0.090}=1.852 \)

    \( \displaystyle \text{D, }\ce{K2SO4}:\quad i\,m=3\times\frac{10/174}{0.090}=1.916 \)

The larger i × m, the higher the boiling point.

Note: The same/similar question has been asked in different years by NTA as per the details given below:

20262025202420232022
26-05 S2
14-05 S2
––––

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