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 |
Isotonic solutions have the same:
1. Vapour pressure
2. Freezing temperature
3. Osmotic pressure
4. Boiling temperature
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 |
During osmosis, the flow of water through a semi-permeable membrane is:
1. | From a solution having higher concentration only. |
2. | From both sides of the semi-permeable membrane with equal flow rates. |
3. | From both sides of the semi-permeable membrane with unequal flow rates. |
4. | From a solution having lower concentration only. |
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.
From the colligative properties of a solution, which one is the best method for the determination of molecular weight of proteins & polymers:
1. Osmotic pressure
2. Lowering of vapour pressure
3. Lowering of freezing point
4. Elevation in boiling point
1 % (w/w) solution of a compound is isotonic with 5 % (w/w) sucrose (sugar) solution. The molecular weight of the compound will be:
1. | 32.4 | 2. | 68.4 |
3. | 129.6 | 4. | 34.2 |