Select Question Set:
filter

On dissolving sugar in water at room temperature solution feels cool to touch. Dissolution of sugar will be most rapid under the following case is:

1.  Sugar crystals in cold water
2.  Sugar crystals in hot water
3.  Powdered sugar in cold water
4.  Powdered sugar in hot water

Subtopic:  Concentration Terms & Henry's Law |
 66%
Level 2: 60%+
Hints

Colligative properties are dependent on which of the following factors?

1. The nature of the solute particles dissolved in the solution.
2. The number of solute particles in the solution.
3. The physical properties of the solute particles dissolved in the solution.
4. The nature of solvent particles.
Subtopic:  Introduction & Colligative properties |
 80%
Level 1: 80%+
Hints

A solution that has the highest boiling point among the following is:

1. 1.0 M NaOH

2. 1.0 M Na2SO4 

3. 1.0 M NH4NO3

4. 1.0 M KNO3

Subtopic:  Elevation of Boiling Point |
 83%
Level 1: 80%+
Hints
Links

advertisementadvertisement

If 8 g of a non-electrolyte solute is dissolved in 114 g of n-octane to reduce its vapor pressure to 80 %, the molar mass (in g mol–1) of the solute is:

[Molar mass of n-octane is 114 g mol–1]

1. 40 2. 60
3. 80 4. 20
Subtopic:  Relative Lowering of Vapour Pressure |
 62%
Level 2: 60%+
NEET - 2020
Hints

Isotonic solutions have the same:

1. Vapour pressure

2. Freezing temperature

3. Osmotic pressure

4. Boiling temperature

Subtopic:  Osmosis & Osmotic Pressure |
 91%
Level 1: 80%+
NEET - 2020
Hints
Links

A 0.1 molal aqueous solution of a weak acid (HA) is 30 % ionized. If Kf for water is 1.86 °C/m, the freezing point of the solution will be:

1. –0.24 °C  2. –0.18 °C
3. –0.54 °C  4. –0.36 °C
Subtopic:  Depression of Freezing Point |
 69%
Level 2: 60%+
AIPMT - 2011
Hints

advertisementadvertisement

200 mL of an aqueous solution contains 1.26 g of protein. The osmotic pressure of this solution at 300 K is found to be 2.57 × 10–3 bar. The molar mass of protein will be:

(R = 0.083 L bar mol–1 K–1):

1. 61038 g mol–1  2. 51022 g mol–1
3. 122044 g mol–1  4. 31011 g mol–1
Subtopic:  Osmosis & Osmotic Pressure |
 72%
Level 2: 60%+
AIPMT - 2011
Hints

The positive deviations from Raoult’s law mean the vapour pressure is:

1. Higher than expected.
2. Lower than expected.
3. As expected.
4. None of the above

Subtopic:  Raoult's Law |
 86%
Level 1: 80%+
Hints
Links

premium feature crown icon
Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly

The type of inter-molecular interactions present in:

(a) n-Hexane and n-octane (i) Van der Waal’s forces of attraction
(b) NaClO4 and water (ii) Ion-dipole interaction
(iii) Dipole-dipole interaction
 
(a) (b)
1. (i) (ii)
2. (ii) (ii)
3. (i) (iii)
4. (iii) (iii)
Subtopic:  Azeotrope |
 65%
Level 2: 60%+
Hints

advertisementadvertisement

Henry’s law constant for the solution of methane in benzene at 298 K is 4.27 × 105 mm Hg. The mole fraction of methane in benzene at 298 K under 760 mm Hg will be:

1. 1.85 × 105

2. 192 × 104

3. 178 × 105

4. 18.7 × 10–5

Subtopic:  Concentration Terms & Henry's Law |
 75%
Level 2: 60%+
Hints
Links

Select Question Set:
filter