- 1(current)
Q. No.Isotonic solutions have the same:
1. Vapour pressure
2. Freezing temperature
3. Osmotic pressure
4. Boiling temperature
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 pressures of these solutions is:
| 1. | P2 > P3 > P1 | 2. | P3 > P1 > P2 |
| 3. | P2 > P1 > P3 | 4. | P1 > P2 > P3 |
The relationship between osmotic pressure at 273 K, when 10 gm glucose (P1), 10 gm urea (P2), and 10 gm sucrose (P3) are dissolved in 250 ml of water, is-
1. P1 > P2 > P3
2. P3 > P1 > P2
3. P2 > P1 > P3
4. P2 > P3 > P1
| 1. | 310°C | 2. | 25.73°C |
| 3. | 12.05 °C | 4. | 37°C |
The osmotic pressure of 5 % (mass-volume) solution of cane sugar at 150 °C (mol. mass of sugar = 342 g/mole) is:
| 1. | 4 atm | 2. | 5.07 atm |
| 3. | 3.55 atm | 4. | 2.45 atm |
1. 62 g/L
2. 0.58 g/L
3. 0.62 g/L
4. 58 g/L
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 |
- 1(current)
Q. No.
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