Relative Lowering of Vapour Pressure

Q.

At 300K, when a solute is added to a solvent its vapour pressure over mercury reduces from 50 mm to 45 mm. The value of mole fraction of solute will be:

1. 0.005
2. 0.010
3. 0.100
4. 0.900

Q. A current of dry air is passed through a bulb containing $5\text{g}$ of a solute in $100\text{g}$ of water and then through water alone. The loss in weight of the solution and pure water were respectively $0.78\text{g}$ and $0.02\text{g}$. Calculate the relative lowering of vapour pressure.

1. 2.04
2. 1.05
3. 0.025
4. 0.09

Q. Dry air was passed through a beaker solution containing $15\text{g}$ of solute and $120\text{g}$ of water. Then, it passed through a beaker containing pure water solvent. Loss in weight of the solution beaker was $2\text{g}$ and loss in weight of solvent beaker was $0.05\text{g}$. What is the molecular weight of the solute ?

1. $90\text{g/mol}$
2. $100\text{g/mol}$
3. $95\text{g/mol}$
4. $85\text{g/mol}$

Q.

If P⁰ and P are the vapour pressure of a solvent and its solution respectively and N₁ and N₂ are the mole fractions of the solvent and solute respectively, then correct relation is

1. 

2. 

3. 

4. 

Q. Calculate the mass of urea required to be dissolved in $100g$ of water, so as the vapour pressure of the solution decreases by $50\%$.
Given: $\text{Molecular mass of urea is}60gmo{l}^{-1}$

1. $33.33g$
2. $22.22g$
3. $222.22g$
4. $333.33g$

Q. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass $78gmo{l}^{-1})$. The vapour pressure of the solution then is 0.845 bar. What is the molar mass of the solid substance?

1. $170gmo{l}^{-1}$
2. $270gmo{l}^{-1}$
3. $370gmo{l}^{-1}$
4. $140gmo{l}^{-1}$

Q. A current of dry air was bubbled through a bulb containing $26.66g$ of an organic compound in $200g$ of water, then through a bulb at the same temperature, containing water and finally through a tube containing anhydrous calcium chloride. The loss of mass in bulb containing water was $0.1g$
and gain in mass of the calcium chloride tube was $2.0g$.
Calculate the molecular mass of the organic substance.

Q. The vapour pressure of pure benzene at ${50}^{o}C\text{is}268mm\text{of}Hg$. How many moles of non-volatile solute per mole of benzene are required to prepare a solution of benzene having a vapour pressure $167mm\text{of}Hg$ at ${50}^{o}C$ ?

Q. The vapour pressure of pure benzene at ${50}^{o}C\text{is}268mm\text{of}Hg$. How many moles of non-volatile solute per mole of benzene are required to prepare a solution of benzene having a vapour pressure $167mm\text{of}Hg$ at ${50}^{o}C$ ?

Q. The vapour pressure of a solution of a non-volatile electrolyte $\text{a}$. in a solvent $\text{b}$. is 95% of the vapour pressure of the solvent at the same temperature. If $M_B=0.3M_A$, where $M_B$ and $M_A$ are molar masses of $B$ and $A$ respectively, the mass ratio of the solvent and the solute are:

1. 0.015
2. 5.7
3. 0.2
4. 4

Q. The vapour Pressure of acetone at ${20}^{\circ }C$ is 185 torr. When 1.2 g of a non- volatile substance was dissolved in 100g of acetone at ${20}^{\circ }C$, its vapour pressure was 183 Torr. The molar mass of the substance is

1. 32
2. 64
3. 128
4. 488

Q.

Vapour pressure of CCl₄ at 25⁰C is 143mm of Hg. 0.5 gm of a non-volatile solute (mol. wt. = 65) is dissolved in 100ml CCl₄. Find the vapour pressure of the solution (Density of CCl₄=1.58g/cm²)

1. 141.43 mm

2. 94.39 mm

3. 199.34 mm

4. 143.99 mm

Q.

The boiling point of a solution of Benzoic acid in Benzene is less than expected because

1. Benzoic acid is an organic solute

2. Benzene is a non-polar solvent

3. Benzoic acid gets associated in Benzene

4. Benzoic acid dissociates in Benzene

Q. Lowering of vapour pressure due to a solute in 1 molal aqueous solution at ${100}^{\circ }C$ is

1. 13.44 mm Hg
2. 14.12 mm Hg
3. 13.2 mm Hg
4. 35.2 mm Hg

Q.

The vapour pressure of acetone at ${20}^{\circ }C$ is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at ${20}^{\circ }C$, its vapour pressure was 183 torr. The molar mass $\left(gmo{l}^{-1}\right)$ of the substance is:

1. 32

2. 64

3. 128

4. 488

Q. The vapour pressure of a solution of a non-volatile electrolyte (A) in a solvent (B) is 95% of the vapour pressure of the solvent at the same temperature. If $M_B=0.3M_A,$ where $M_A$ and $M_B$ are molecular weights of (A) and (B) respectively, then the weight ratio of the solvent and solute are

1. 69.8
2. 66.6
3. 5.7
4. 76.6

Q.

Pure benzene freezes at 5.45°C at a certain places but a 0.374 m solution of tetrachloroethane In benzene freezes at 3.55°C. The $K_f$ for benzene is

1. 5.08 k kg mol^-1
2. 508 K Kg mol^-1
3. 0.508 K kg mol ^-1
4. 50.8°C kg mol^-1

Q.

The boiling point of a solution of Benzoic acid in Benzene is less than expected because

1. Benzoic acid is an organic solute

2. Benzene is a non-polar solvent

3. Benzoic acid gets associated in Benzene

4. Benzoic acid dissociates in Benzene

Q. Mass of non-volatile solute (urea) needs to be dissolved in 100 g of water in order to decrease the vapour pressure of water by 30%. What will be the molality of solution?

1. 23.88 m
2. 28.88 m
3. 25.88 m
4. 35.88 m

Q. The vapour pressure of acetone at ${20}^{\circ }$$C$ is 185 $torr$. When 1.2 $g$ of a non-volatile substance was dissolved in 100 $g$ of acetone at ${20}^{\circ }$$C$, its vapour pressure was 183 $torr$. The molar mass $\left(g/mol\right)$ of the substance is:

1. 32
2. 64
3. 128
4. 488

Q.

The vapour pressure of pure benzene $C_6{H}_6$ at ${50}^{\circ }C$ is 268 torr. How many moles on non-volatile solute per mole of benzene is required to prepare a solution of benzene having a vapour pressure of 167 torr at ${50}^{\circ }C$?

1. 0.377

2. 0.605

3. 0.623

4. 0.395
   

Q. 4.0 g of a substance when dissolved in $36.0\text{g}$ of water lowered its vapour pressure by $0.5\text{mm Hg}$ at a given temperature. The vapour pressure of water at this temperature is $25.0\text{mm Hg}$. Calculate the molecular weight of solute.

1. $100\text{g/mol}$
2. $100\text{g/mol}$
3. $98\text{g/mol}$
4. $100\text{g/mol}$

Q.

The freezing point of 1 m NaCl solution, assuming NaCl to be 100% dissociated in water

1. -1.86°C

2. -3.72°C

3. +1.86°C

4. +3.72°C

Q. The vapour pressure of a solution of a non-volatile electrolyte (A) in a solvent (B) is 95% of the vapour pressure of the solvent at the same temperature. If $M_B=0.3M_A,$ where $M_A$ and $M_B$ are molecular weights of (A) and (B) respectively, then the weight ratio of the solvent and solute are

1. 69.8
2. 66.6
3. 5.7
4. 76.6

Q.

The solubility of Iodine in pure water is less than its solubility in potassium iodide solution, because

1. Iodine dissociates in water

2. Iodine reacts with water

3. Iodine does not react with water

4. Cannot be predicted

Q. Calculate the vapour pressure lowering caused by addition of $50\text{g}$ of sucrose $\left(\text{molecular mass = 342}\right)$ to $500\text{g}$ water, if the vapour pressure of pure water at ${25}^{o}C$ is $23.8\text{mm Hg}$.

1. $0.94\text{mm Hg}$
2. $0.42\text{mm Hg}$
3. $1.24\text{mm Hg}$
4. $0.124\text{mm Hg}$

Q.

Vapour pressure of a solution of 5g of non-electrolyte in 100g of water at a particular temperature is 2985 N/m². The vapour pressure of pure water is 3000 N/m². The molecular weight of the solute is:

1. 60

2. 120

3. 180

4. 380

Q. Consider the following cases A and B in Ostwald-Walker Method:
Case A : Dry air is passed through a pure solvent and then, a dehydrating agent.

Case B: Dry air is passed through a pure solvent and then a solution with non-volatile solute and then, a dehydrating agent

Here,
$p^{\circ }\text{is the vapour pressure of the pure solvent}$
$p_s\text{is the vapour pressure of the solution}$

Choose the correct option(s) :

1. $\text{In case B, loss of weight of solvent container}\propto P_S$
2. $\text{In case A, loss of weight of solvent container is}\propto P^{\circ }$
3. $\text{In case B, the gain in weight of solution}\propto (P^{\circ }-P_S)$
4. $\text{In case A, gain in weight of dehydrating agent}\propto P^{\circ }$

Q.

The freezing point of 1 m NaCl solution, assuming NaCl to be 100% dissociated in water

1. -1.86°C

2. -3.72°C

3. +1.86°C

4. +3.72°C

Q. The vapour pressure of dilute aqueous solution of glucose is 750 mm of Hg at 373 K. The mole fraction of glucose is :

1. $\frac{1}{10}$
   
2. $\frac{1}{7.6}$
   
3. $\frac{1}{35}$
4. $\frac{1}{76}$