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 |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
Type of solution | Example |
a. Solid in gas | i. Aerated water |
b. Gas in liquid | ii. Smoke |
c. Liquid in solid | iii. Solution of hydrogen in palladium |
d. Gas in solid | iv. Amalgams |
a | b | c | d | |
1. | i | iii | iv | ii |
2. | ii | i | iv | iii |
3. | iii | i | iv | ii |
4. | iv | i | ii | iii |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
Colligative properties depend on:
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. |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
The solubility of a gas in a liquid decreases with an increase in temperature because:
1. | Dissolution of a gas in a liquid is an endothermic process. |
2. | Dissolution of a gas in a liquid is an exothermic process. |
3. | Gases are highly compressible. |
4. | All of the above. |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
An aqueous solution of hydrochloric acid -
1. Obeys Raoult's law
2. Shows negative deviations from Raoult's law
3. Shows positive deviations from Raoult's law
4. Obeys Henry's law at all compositions
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
Consider the following statements about the composition of the vapour over an ideal 1:1 molar mixture of benzene and toluene. The correct statement is:
Assume that the temperature is constant at 25 oC.
(Given, vapour pressure data at 25 °C, benzene = 12.8 kPa, toluene = 3.85 kPa)
1. | The vapour will contain a higher percentage of toluene. |
2. | The vapour will contain equal amounts of benzene and toluene. |
3. | Not enough information is given to make a prediction. |
4. | The vapour will contain a higher percentage of benzene. |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
To minimise the painful effects accompanying deep sea diving, oxygen diluted with less soluble helium gas is used as breathing gas by the divers. This is an example of the application of -
1. | Raoult's law | 2. | Henry's law |
3. | Ideal gas Equation | 4. | All of the above |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
The graph representing Henry's law is given below:
The value of the slope will be
1. | \(4\times10^{-4}~\mathrm{torr}\) | 2. | \(5\times10^{-4}~\mathrm{torr}\) |
3. | \(4\times10^{4}~\mathrm{torr}\) | 4. | \(5\times10^{4}~\mathrm{torr}\) |
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.
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 x 10-5
2. 192 x 10-4
3. 178 x 10-5
4. 18.7 x 10-5
To unlock all the explanations of this course, you need to be enrolled.
To unlock all the explanations of this course, you need to be enrolled.