(A) The relative lowering of vapour pressure is given by $\frac{P^o - P}{P^o} = \chi_{solute} = \frac{n}{n + N}$.Since the amount of solvent (water) i

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Same as that of $2^{nd}$ solution

(A) The relative lowering of vapour pressure is given by $\frac{P^o - P}{P^o} = \chi_{solute} = \frac{n}{n + N}$.Since the amount of solvent (water) is the same $(1 \ L)$ and the amount of solute is small,the mole fraction is directly proportional to the number of moles of solute $(n)$.For the first solution,moles of urea $= \frac{12 \ g}{60 \ g/mol} = 0.2 \ mol$.For the second solution,moles of cane sugar $= \frac{68.4 \ g}{342 \ g/mol} = 0.2 \ mol$.Since the number of moles of solute is the same in both solutions,the lowering of vapour pressure will be the same.

Class 12 Chemistry Solutions Lowering of vapour pressure

Medium

$A$ solution is obtained by dissolving $12 \ g$ of urea (mol.wt. $60$) in a litre of water. Another solution is obtained by dissolving $68.4 \ g$ of cane sugar (mol.wt. $342$) in a litre of water at the same temperature. The lowering of vapour pressure in the first solution is

A
Same as that of $2^{nd}$ solution
B
Nearly one-fifth of the $2^{nd}$ solution
C
Double that of $2^{nd}$ solution
D
Nearly five times that of $2^{nd}$ solution

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Lowering of vapour pressure

Assertion : If a liquid solute more volatile than the solvent is added to the solvent,the vapour pressure of the solution may increase i.e.,$p_s > p^o$.
Reason : In the presence of a more volatile liquid solute,only the solute will form the vapours and solvent will not.

DifficultAIIMS 2016

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The vapour pressure of pure $CHCl_3$ and $CH_2Cl_2$ are $200 \,atm$ and $41.5 \,atm$ respectively. The weights of $CHCl_3$ and $CH_2Cl_2$ are respectively $11.9 \,g$ and $17 \,g$. The vapour pressure of the solution will be (in $,atm$)

DifficultAIIMS 2019

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At $298 \ K$,the vapour pressure of a solution of $7.5 \ g$ of non-volatile solute in $90 \ g$ of water is $2.8 \ kPa$. If $18 \ g$ of water is added to this solution,the vapour pressure becomes $2.81 \ kPa$ at the same temperature. The molar mass of the solute in $g \ mol^{-1}$ is:

DifficultAP EAMCET 2019

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$A$ solution of a non-volatile solute is obtained by dissolving $2 \ g$ of solute in $50 \ g$ of benzene. Calculate the vapour pressure of the solution if the vapour pressure of pure benzene is $640 \ mmHg$ at $25^{\circ} C$. [Molar mass of benzene $= 78 \ g \ mol^{-1}$,Molar mass of solute $= 64 \ g \ mol^{-1}$] (in $mm \ Hg$)

DifficultMHT CET 2024

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At $50^{\circ} C$,the vapour pressure of pure benzene is $268 \ torr$. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of $167 \ torr$ at the same temperature is (molar mass of benzene $= 78 \ g \ mol^{-1}$)

MediumTS EAMCET 2023

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$A$ solution is obtained by dissolving $12 \ g$ of urea (mol.wt. $60$) in a litre of water. Another solution is obtained by dissolving $68.4 \ g$ of cane sugar (mol.wt. $342$) in a litre of water at the same temperature. The lowering of vapour pressure in the first solution is

Similar Questions

Assertion : If a liquid solute more volatile than the solvent is added to the solvent,the vapour pressure of the solution may increase i.e.,$p_s > p^o$.Reason : In the presence of a more volatile liquid solute,only the solute will form the vapours and solvent will not.

The vapour pressure of pure $CHCl_3$ and $CH_2Cl_2$ are $200 \,atm$ and $41.5 \,atm$ respectively. The weights of $CHCl_3$ and $CH_2Cl_2$ are respectively $11.9 \,g$ and $17 \,g$. The vapour pressure of the solution will be (in $,atm$)

At $298 \ K$,the vapour pressure of a solution of $7.5 \ g$ of non-volatile solute in $90 \ g$ of water is $2.8 \ kPa$. If $18 \ g$ of water is added to this solution,the vapour pressure becomes $2.81 \ kPa$ at the same temperature. The molar mass of the solute in $g \ mol^{-1}$ is:

At $50^{\circ} C$,the vapour pressure of pure benzene is $268 \ torr$. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of $167 \ torr$ at the same temperature is (molar mass of benzene $= 78 \ g \ mol^{-1}$)

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Assertion : If a liquid solute more volatile than the solvent is added to the solvent,the vapour pressure of the solution may increase i.e.,$p_s > p^o$.
Reason : In the presence of a more volatile liquid solute,only the solute will form the vapours and solvent will not.