Relative Lowering of Vapour Pressure

Q. 18 g of glucose $\left(C_6{H}_{12}{O}_6\right)$ is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at ${100}^{\circ }C$ is

1. 759.00 torr
2. 7.60 torr
3. 76.00 torr
4. 752.40 torr.

Q. The vapour pressure of an aqueous solution of sugarcane (molecular weight 342) is $756mmHg\text{at}{100}^{\circ }C$. How many grams of sugar are present per $1000g$ of water?
Vapour pressure of pure water at ${100}^{\circ }C$ is $760mmHg$

1. $50g$
2. $77.5g$
3. $100g$
4. $110g$

Q. The vapour pressure of benzene at a certain temperature is 640 mm. of Hg. A non-volatile and non-electrolyte solid, weighing 2.175 g is added to 39.08 g of benzene. If the vapour pressure of the solution is 600 mm of Hg. What is the molecular weight of solid substance?

1. 49.50
2. 59.60
3. 69.40
4. 79.82

Q. From relative lowering of vapour pressure , molecular mass of non-volatile solute can be determined.
Which of the following relations is correct?
Here,
$P_A^{\circ }\text{is vapour pressure of pure solvent}$
$P_S\text{is vapour pressure of solution}$
$w_A,w_B\text{are mass of solvent and solute respectively}$
$M_A,M_B\text{are molecular masses of solvent and solute respectively}$

1. $M_B=\frac{w_B\times M_A\times P_S}{w_A\times (P_A^{\circ }-P_S)}$
2. $M_B=\frac{w_A\times M_A\times P_S}{w_B\times (P_A^{\circ }-P_S)}$
3. $M_B=\frac{w_B\times (P_A^{\circ }-P_S)}{w_A\times M_A\times P_S}$
4. None of the above

Q. The vapour pressure of pure benzene at certain temperature is $640mm$ of $Hg$. A non-volatile non-electrolyte solid weighing $2.175g$ is added to $39.0g$ of benzene. The vapour pressure of the solution is $600mm$ of $Hg$. What is the molar mass of the solid substance?

1. $125g$
2. $105g$
3. $80.5g$
4. $65.25g$

Q. Calculate the relative lowering of vapour pressure when $12g$ urea (Mol weight 60 g/mol) is dissolved in $90g$ of water.

1. $0.011$
2. $0.11$
3. $0.038$
4. $0.38$

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

Q. In Ostwald-Walker method, dry air is passed through a solution beaker, then a pure solvent beaker and then, through the dehydrating agent. Which of the following statement(s) is/are correct regarding the above setup?
Here,
$p^{\circ }\text{is the vapour pressure of the pure solvent}$
$p_s\text{is the vapour pressure of the solution}$

1. $\text{Loss in weight of solvent beaker}\propto p^{\circ }-p_s$
2. $\text{Gain in weight of dehydrating agent}\propto p^{\circ }$
3. $\text{Loss in weight of solvent beaker}\propto p^{\circ }$
4. $\text{Gain in weight of solution beaker}\propto p^{\circ }-p_s$

Q. At ${20}^{\circ }C$ The vapour pressure of pure ether is $442mmHg$. When $6.1g$ of a non volatile substance is dissolved in $50g$ of ether (molecular weight 74). The vapour pressure of solution falls to $410mmHg$.
What is the molecular weight of the substance?
Assume the solution to be non-dilute.

1. $74.2g$
2. $33.9g$
3. $115.6g$
4. $50.5g$

Q. The vapour pressure of benzene at ${80}^{\circ }\text{C}$ is lowered by $10\text{mm Hg}$ by dissolving $2\text{g}$ of a nonvolatile substance in $78\text{g}$ of benzene. The vapour pressure of pure benzene at ${80}^{\circ }\text{C}$ is $750\text{mm Hg}$. The molecular mass of the substance will be:

1. $15$
2. $1500$
3. $14.8$
4. $148$

Q. The vapour pressure of $H_{2}O$ is $23.8\text{mm Hg}$ at ${25}^{o}C$. What is the vapour pressure of a solution of $28.5\text{g}$ of sucrose, $C_{12}{H}_{22}{O}_{11}\left(\text{Molecular mass= 342 u}\right)$ in $100\text{g}$ of $H_{2}O$?

1. $23.1\text{mm Hg}$
2. $32.5\text{mm Hg}$
3. $20.6\text{mm Hg}$
4. $20.00\text{mm Hg}$

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.

1. $m=50g/mol$
2. $m=64.4g/mol$
3. $m=56g/mol$
4. $m=45.6g/mol$

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. A non-volatile solute (urea) is dissolved in $100g$ of water in order to decrease the vapour pressure of water by $25\%$. Which of the following option(s) is/are correct?
Assume the solution to be non-dilute.

1. $\text{Mass of urea added is}111.1g$
2. $\text{Mass of urea added is}88.1g$
3. $\text{Molality of the solution is}18.52m$
4. $\text{Molality of the solution is}14.52m$

Q. Dry air was bubbled through a bulb containing $22g$ of an non-volatile organic compound in $150g$ of water, then it passed through a bulb containing water at the same temperature and finally through a bulb containing anhydrous calcium chloride. The weight loss in bulb containing water was $0.075g$
and gain in mass of the calcium chloride tube was $1.5g$.
Calculate the molecular mass of the organic substance.

1. $42.16g/mol$
2. $50.16g/mol$
3. $59.16g/mol$
4. $64.16g/mol$

Q. Dry air was passed through a solution containing $20g$ of a substance in $100.0g$ of water and then through pure water. The loss in mass of the solution was $3g$ and that of pure water was $0.06g$. Calculate the molar mass of the substance.

1. $198.2g/mol$
2. $180g/mol$
3. $108.2g/mol$
4. $208.6g/mol$

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 ${50}^{o}C$ ?

1. $0.60$
2. $0.25$
3. $0.80$
4. $0.32$

Q. Which of the following can be measured by the Ostwald-Walker dynamic method ?

1. Relative lowering of vapour pressure.
2. Depression in freezing point
3. Ebullioscopic constant
4. All of the above.

Q. When $3g$ urea (Mol weight 60 g/mol) is dissolved in $45g$ of water. The relative lowering of vapour pressure is:

1. $0.18$
2. $0.12$
3. $0.08$
4. $0.02$

Q. 0.5 g of a nonvolatile organic compound (molecular weight 65) is dissolved in $100mLofCC{l}_4$. If the vapour pressure of pure $CC{l}_{4}is140mmHg$, then calculate the vapour pressure of the solution.
$\text{(Density of}CC{l}_4\text{solution is}1.6gm{L}^{-1}$

1. $145.5mmHg$
2. $138.9mmHg$
3. $165.2mmHg$
4. $94.7mmHg$

Q. A solution containing $30\text{g}$ of a non-volatile solute in exactly $90\text{g}$ of water has a vapour pressure of $21.85\text{mm of Hg}$ at ${25}^{o}C.$ Further, $18\text{g}$ of water is then added to the solution, the new vapour pressure becomes $22.15\text{mm of Hg}$ at ${25}^{o}C.$ Calculate:
(i) Molar mass of the solute (in g/mol) and
(ii) Vapour pressure of water at ${25}^{o}C.$

1. $\left(i\right)=23.87,\left(ii\right)=97\text{mm Hg}$
2. $\left(i\right)=87,\left(ii\right)=67.9\text{mm Hg}$
3. $\left(i\right)=67.9,\left(ii\right)=23.87\text{mm Hg}$
4. $\left(i\right)=23.87,\left(ii\right)=67.9\text{mm Hg}$

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

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. $95g/mol$
2. $100g/mol$
3. $90g/mol$
4. $85g/mol$

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 incorrect option:

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

Q. Which of the following is correct regarding vapour pressure of the solution formed when a non-volatile solute (solid) is added in a pure volatile solvent?

1. Vapour pressure of the solution increases.
2. Vapour pressure of the solution decreases.
3. Vapour pressure of the solution remains the same.
4. Vapour pressure of the solution may increase or decrease.

Q. In Ostwald-Walker method, dry air is passed through a series of beaker containing solution and solvent and finally through a dehydrating agent as shown in the figure


Which of the following is incorrect regarding the above setup?
Here,
$p^{\circ }\text{is the vapour pressure of the pure solvent}$
$p_s\text{is the vapour pressure of the solution}$

1. Weight of beaker 3 increase
2. Weight of beaker 2 increases
3. Weight of beaker 4 decreases
4. $\text{Loss in weight of beaker 1}\propto p_s$

Q. If the relative lowering of pressure of o-xylene is 0.005 due to addition of 0.5 grams of non-volatile solute in 500 grams of solvent, what is the molecular weight of the solute?

1. 21.3 g/ mol
2. 23.1 g/mol
3. 32.1 g/mol
4. 12.3 g/mol

Q.

The boiling points of three saturated hydrocarbons A, B and C are $-162°C,-42.2°C$ and$-0.5°C$ respectively. Which of these three hydrocarbons would have the maximum number of carbon atoms in its molecule?

Q. Calculate the vapour pressure of an aqueous solution of $1.0$ molal glucose solution at ${100}^{o}C.$

1. $746.32\text{mm Hg}$
2. $760\text{mm Hg}$
3. $13.68\text{mm Hg}$
4. $136.8\text{mm Hg}$

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. $140gmo{l}^{-1}$