Q. No.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.
Which of the following mixtures forms the maximum boiling azeotrope?
| 1. | Heptane + Octane | 2. | Water + Nitric acid |
| 3. | Ethanol + Water | 4. | Acetone + Carbon disulphide |
freezes at 5.5C. At what temperature will a solution of 10.44 g of in 200 g of freeze ? Kf () = 5.12C/m
1. 4.60C
2. 0.89C
3. 5.50C
4. None of the above
A solution containing 10 g/dm3 of urea (molecular mass = 60 g mol-1) is isotonic with a 5 % solution of a non-volatile solute. The molecular mass of this non-volatile solute is:
| 1. | 25 g mol-1 | 2. | 300 g mol-1 |
| 3. | 350 g mol-1 | 4. | 200 g mol-1 |
The freezing point depression constant for water is 1.86 oC m-1. If 5.00 g Na2SO4 is dissolved in 45.0 g H2O, the freezing point is changed by -3.82 oC. The Van’t Hoff factor for Na2SO4 is:
| 1. | 2.63 | 2. | 3.11 |
| 3. | 0.381 | 4. | 2.05 |
The largest freezing point depression among the following 0.10 m solutions is shown by:
| 1. | \(\mathrm{KCl}\) | 2. | \(\mathrm{C_6H_{12}O_6}\) |
| 3. | \(\mathrm{Al}_2(\mathrm{SO_4})_3\) | 4. | \(\mathrm{K_2SO_4}\) |
The boiling point of 0.2 mol kg–1 solution of X in water is greater than the equimolal solution of Y in water. The correct statement in this case is:
| 1. | X is undergoing dissociation in water. |
| 2. | Molecular mass of X is greater than the molecular mass of Y. |
| 3. | Molecular mass of X is less than the molecular mass of Y. |
| 4. | Y is undergoing dissociation in water while X undergoes no change. |
1. \(\Delta H_\text{mix}\)
2. \(\Delta S_\text{mix}\)
3. \(\Delta V_\text{mix}\)
4. \(\Delta P=P_{\text{observed}}-P_{\text{Raoult}}\)
Liquid ↔ Vapour
Which of the following relation is correct?
1. \(\frac{dlnp}{dT}=\frac{-\Delta H_v}{RT}\)
2. \(\frac{dlnp}{dT^2}=\frac{-\Delta H_v}{T^2}\)
3. \(\frac{dlnp}{dT}=\frac{\Delta H_v}{RT^2}\)
4. \(\frac{dlnG}{dT^2}=\frac{-\Delta H_v}{RT^2}\)
The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 formed from air dissolved in 10 moles of water at 298 K and 5 atm pressure is-
| 1. | 4.0 × 10–4 mol | 2. | 4.0 × 10–5 mol |
| 3. | 5.0 × 10–4 mol | 4. | 4.0 × 10–6 mol |
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