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Solutions · Question 34

JEE Main 2025 (Online) 2nd April Morning ShiftMCQMedium📈 Topic in 10% papers+4−1
1

Which of the following graph correctly represents the plots of KH at 1 bar for gases in water versus temperature?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]

Answer

(2) [figure]

Solution

Henry's law: p = KH x. A larger KH means a lower solubility of the gas.

As per NCERT (Table 1.2), KH at 293 K: He = 144.97 kbar, N2 = 76.48 kbar, O2 = 34.86 kbar, and CH4 (298 K) = 0.413 105 bar.

Order of KH: He > N2 > CH4 > O2; so He must lie highest and O2 lowest. Graphs (1) and (4) put O2 on top rejected.

Solubility of gases decreases with rise in temperature KH first increases with temperature; for gases in water it passes through a maximum and then decreases at higher temperatures, giving bell-shaped curves.

Graph (3) shows KH first decreasing, which is wrong.

Graph (2) (He > N2 > CH4, each rising to a maximum and then falling) is correct.

Answer: (2)

JEE Main 2025 (Online) 2nd April Morning Shift

Keep practising — Solutions

2

When 1 g each of compounds AB and AB2 are dissolved in 15 g of water separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. The atomic mass of A (in amu) is 10-1 (Nearest integer) (Given : Molal boiling point elevation constant is 0.5 K kg mol-1)

JEE Main 2025 (Online) 2nd April Evening Shift
3

1.24 g of AX2 (molar mass 124 g mol-1) is dissolved in 1 kg of water to form a solution with boiling point of 100.0156°C, while 25.4 g of AY2 (molar mass 250 g mol-1) in 2 kg of water constitutes a solution with a boiling point of 100.0260°C. Kb(H2O) = 0.52 kg mol-1 Which of the following is correct?

  1. AX2 and AY2 (both) are completely unionised.
  2. AX2 and AY2 (both) are fully ionised.
  3. AX2 is completely unionised while AY2 is fully ionised.
  4. AX2 is fully ionised while AY2 is completely unionised.
JEE Main 2025 (Online) 29th January Morning Shift
4

If A2B is 30% ionised in an aqueous solution, then the value of van't Hoff factor (i) is 10-1.

JEE Main 2025 (Online) 29th January Morning Shift
5

What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non-volatile solute (molar mass 256 g mol-1) and the decrease in freezing point is 0.40 K?

  1. 4.43 K kg mol-1
  2. 3.72 K kg mol-1
  3. 5.12 K kg mol-1
  4. 1.86 K kg mol-1
JEE Main 2025 (Online) 28th January Morning Shift
6

Assume a living cell with 0.9% (w/w) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data upto first decimal place only) The cell will :

  1. shrink since solution is 0.45% (w/w) as a result of association of glucose molecules (due to hydrogen bonding)
  2. shrink since solution is 0.5% (w/w)
  3. show no change in volume since solution is 0.9% (w/w)
  4. swell up since solution is 1% (w/w)
JEE Main 2025 (Online) 28th January Evening Shift

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