The mole fraction of ethylene glycol (C2H6O2) in a solution containing 20% of C2H6O2 by mass is-

1. 0.932

2. 0.0672

3. 0.832

4. 0.732

Subtopic:  Concentration Terms & Henry's Law |
 71%
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If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would dissolve in 1 litre of water?
(
Partial pressure of nitrogen = 0.987 bar, Henry’s law constant for N2 at 293 K = 76.48 kbar).

1. 0.716 mmol

2. 0.616 mmol

3. 0.726 mmol

4. 0.516 mmol

Subtopic:  Concentration Terms & Henry's Law |
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The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. The molar mass of the solute is-

Kb for benzene is 2.53 K kg mol–1

1. 55 g mol–1
2. 64 g mol–1
3. 58 g mol–1  
4.
68 g mol–1

Subtopic:  Elevation of Boiling Point |
 64%
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45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. The freezing point of the solution is -

1. 268.15 K
2. 270.95 K
3. 272.75 K
4. 267.15 K

Subtopic:  Depression of Freezing Point |
 64%
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The mole fraction of glucose (C6H12O6) in an aqueous binary solution is 0.1. The mass percentage of water in it to the nearest integer is-
1. 44%
2. 49 %
3. 47 %
4. 41 %

Subtopic:  Concentration Terms & Henry's Law |
 62%
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On mixing, heptane, and octane form an ideal solution at 373 K, the vapor pressures of the two liquid components (Heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be:
(molar mass of heptane = 100 g mol−1 and of octane =114 g 1 mol−1)

1. 144.5 kPa

2. 72.0 kPa

3. 36.1 kPa

4. 96.2 kPa

Subtopic:  Raoult's Law |
 75%
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The correct option for the value of vapour pressure of a solution at 45 °C with benzene to octane in a 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 gas]
1. 336 mm of Hg
2. 350 mm of Hg
3. 160 mm of Hg
4. 168 mm of Hg

Subtopic:  Raoult's Law |
 72%
From NCERT
NEET - 2021
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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 decreasing order of osmotic the pressure of these solutions is:

1. P2 > P3 > P1 2. P3 > P1 > P2
3. P2 > P1 > P3 4. P1 > P2 > P3
Subtopic:  Osmosis & Osmotic Pressure |
 69%
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NEET - 2021
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Choose a false statement among the following:

1.  Raoult's law states that the vapour pressure of a component over a solution is proportional to its mole fraction.
2.  The osmotic pressure (\(\pi\)) of a solution is given by the equation \(\pi\)=MRT, where M is the molarity of the solution.
3.  The correct order of osmotic pressure for 0.10 M aqueous solution of each compound is BaCl2 > KCl > CH3COOH > sucrose.
4.  Two sucrose solutions of the same molarity prepared in different solvents will have the same depression in the freezing point. 

Subtopic:  Depression of Freezing Point | Raoult's Law |
 73%
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A solution of urea (molar mass 56 g mol–1) boils at 100.18 ºC at atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol–1 respectively, the above solution will freeze at:

1. –6.54 ºC

2. –0.654 ºC

3. 6.54 ºC

4. 0.654 ºC

Subtopic:  Elevation of Boiling Point | Depression of Freezing Point |
 75%
From NCERT
AIPMT - 2005
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