1. | 1000 g of solvent | 2. | 500 mL of solvent |
3. | 500 g of solvent | 4. | 100 mL of solvent |
One mole of sugar is dissolved in three moles of water at 298 K. The relative lowering of vapour pressure is:
1. | 0.25 | 2. | 0.15 |
3. | 0.50 | 4. | 0.33 |
gas | KH/k bar |
Ar | 40.3 |
CO2 | 1.67 |
HCHO | 1.83 × 10–5 |
CH4 | 0.413 |
1. | Ar < CO2 < CH4 < HCHO |
2. | Ar < CH4 < CO2 < HCHO |
3. | HCHO < CO2 < CH4 < Ar |
4. | HCHO < CH4 < CO2 < Ar |
Assertion (A): | Helium is used to dilute oxygen in the diving apparatus. |
Reason (R): | Helium has a high solubility in O2. |
1. | (A) is False but (R) is True. |
2. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
3. | Both (A) and (R) are True and (R) is not the correct explanation of (A). |
4. | (A) is True but (R) is False. |
1. | 0.05 M NaCl | 2. | 0.1 M KCl |
3. | 0.1 M MgSO4 | 4. | 1 M NaCl |
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
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 |
Beans get cooked earlier in a pressure cooker, because:
1. | The boiling point increases with increasing pressure |
2. | The boiling point decreases with increasing pressure |
3. | The extra pressure of the pressure cooker softens the beans |
4. | Internal energy is not lost while cooking in a pressure cooker |
The ideal solution indicates:
1. | A – B attraction force is greater than A – A and B – B. |
2. | A – B attraction force is less than A – A and B – B. |
3. | Attraction force remains the same in A – A and B – B. |
4. | The volume of the solution is different from the sum of the volume of the solute and solvent. |
A solution contains a non-volatile solute of molecular mass M2. The molecular mass of solute in terms of osmotic pressure is:
where:
m2 → Mass of solute
V → Volume of solution
\(\pi\) → Osmotic pressure
1.
2.
3.
4.