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Solutions · Question 8

NEET (UG) 2024MCQEasy📈 Topic in 38% papers+4−1
1

The Henry's law constant (KH) values of three gases (A, B, C) in water are 145, 2 10-5 and 35 kbar respectively. The solubility of these gases in water follows the order:

  1. B > A > C
  2. B > C > A
  3. A > C > B
  4. A > B > C

Answer

(2) B > C > A

Solution

Henry's law: p = KH x

x = pKH, so at the same partial pressure, solubility 1KH.

KH: B (2 10-5) < C (35) < A (145)

Solubility: B > C > A

Answer: (2)

NEET (UG) 2024

Keep practising — Solutions

2

The plot of osmotic pressure ( ) vs concentration ( mol L-1) for a solution gives a straight line with slope 25.73 L bar mol-1. The temperature at which the osmotic pressure measurement is done is (Use R = 0.083 L bar mol-1 K-1)

  1. 37
  2. 310
  3. 25.73
  4. 12.05
NEET (UG) 2024
3

Arrange the following compounds in increasing order of their solubilities in chloroform: NaCl, CH3OH, cyclohexane, CH3CN

  1. NaCl < CH3CN < CH3OH < Cyclohexane
  2. CH3OH < CH3CN < NaCl < Cyclohexane
  3. NaCl < CH3OH < CH3CN < Cyclohexane
  4. Cyclohexane < CH3CN < CH3OH < NaCl
RE-NEET (UG) 2024
4

Mass of glucose ( C6H12O6) required to be dissolved to prepare one litre of its solution which is isotonic with 15 g L-1 solution of urea (NH2CONH2) is (Given: Molar mass in g mol-1 C : 12, H : 1, O : 16, N : 14)

  1. 55 g
  2. 15 g
  3. 30 g
  4. 45 g
RE-NEET (UG) 2024
5

KH value for some gases at the same temperature 'T' are given : [figure] where KH is Henry's Law constant in water. The order of their solubility in water is :

  1. HCHO < CH4 < CO2 < Ar
  2. Ar < CO2 < CH4 < HCHO
  3. Ar < CH4 < CO2 < HCHO
  4. HCHO < CO2 < CH4 < Ar
NEET (UG) 2022 (Phase 2)
6

The following solutions were prepared by dissolving 10 g of glucose ( C6H12O6) in 250 ml of water (P1), 10 g of urea (CH4N2O) in 250 ml of water (P2) and 10 g of sucrose ( C12H22O11) in 250 ml of water (P3). The right option for the decreasing order of osmotic pressure of these solutions is :

  1. P3 > P1 > P2
  2. P2 > P1 > P3
  3. P1 > P2 > P3
  4. P2 > P3 > P1
NEET (UG) 2021
7

The correct option for the value of vapour pressure of a solution at 45 C with benzene to octane in molar ratio 3 : 2 is : [At 45 C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume ideal solution]

  1. 350 mm of Hg
  2. 160 mm of Hg
  3. 168 mm of Hg
  4. 336 mm of Hg
NEET (UG) 2021
8

The freezing point depression constant (Kf) of benzene is 5.12 K kg mol-1. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is rounded off upto two decimal places :

  1. 0.80 K
  2. 0.40 K
  3. 0.60 K
  4. 0.20 K
NEET (UG) 2020
9

The mixture which shows positive deviation from Raoult's law is :

  1. Benzene + Toluene
  2. Acetone + Chloroform
  3. Chloroethane + Bromoethane
  4. Ethanol + Acetone
NEET (UG) 2020
10

For an ideal solution, the correct option is :

  1. mix H = 0 at constant T and P
  2. mix G = 0 at constant T and P
  3. mix S = 0 at constant T and P
  4. mix V 0 at constant T and P
NEET (UG) 2019
11

The mixture that forms maximum boiling azeotrope is :

  1. Ethanol + Water
  2. Acetone + Carbon disulphide
  3. Heptane + Octane
  4. Water + Nitric acid
NEET (UG) 2019

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