Answer
64.8k+ views
Hint: We know that mole fraction of solute is directly proportional relative lowering of vapor pressure. So, we have to identify the solute whose mole fraction is greater among the four options.
Complete step-by-step answer:
First, we have to establish the relation between lowering of vapor pressure and mole fraction.
We know that the relation between mole fraction and vapour pressure of solvent and vapor pressure of solution is:
${p_1} = {x_1}{p_1}^0$ …… (1)
Where, ${p_1}$is the vapour pressure of solution, ${x_1}$is mole fraction of solvent and ${p_1}^0$ is vapour pressure of solvent.
The reduction in vapour pressure of solvent $\left( {\Delta {p_1}} \right)$ is given as:
$\Delta {p_1} = {p_1}^0 - {p_1}$
We have to put the value of ${p_1}$ from equation (1) in the below equation.
$\begin{array}{c}\Delta {p_1} = {p_1}^0 - {p_1}^0{x_1}\\ = {p_1}^0\left( {1 - {x_1}} \right)\end{array}$ …… (2)
We know that, ${x_1} + {x_2} = 1$, where ${x_1}$ is mole fraction of solvent and ${x_2}$ is mole fraction of solute.
$\begin{array}{c}{x_1} + {x_2} = 1\\{x_2} = 1 - {x_1}\end{array}$
Now, we have to put the value of $1 - {x_1}$ in equation (2).
$\Delta {p_1} = {x_2}{p_1}^0$ …… (3)
Now, we divide equation (3) by ${p_1}^0$.
$\begin{array}{c}\dfrac{{\Delta {p_1}}}{{{p_1}^0}} = \dfrac{{{x_2}{p_1}^0}}{{{p_1}^0}}\\\dfrac{{\Delta {p_1}}}{{{p_1}^0}} = {x_2}\\\dfrac{{{p_1}^0 - {p_1}}}{{{p_1}^0}} = \dfrac{{{n_2}}}{{{n_2} + {n_1}}}\end{array}$ ……. (4)
So, the relation establishes that lowering of vapor pressure is equal to mole fraction of solute. Mole fraction again related to moles of solute. So, we have to find the solution of highest molality to identify the solution possessing in which lowering of vapour pressure is highest.
In the given question, molality of urea is $0.2\;{\rm{m}}$, molality of glucose is 0.1 m, molality of ${\rm{MgS}}{{\rm{O}}_{\rm{4}}}$ is 0.1 m and molality of ${\rm{BaC}}{{\rm{l}}_{\rm{2}}}$ is 0.1 m.
The highest molality of urea indicates that moles of urea are highest among all the options.
Highest moles of solute indicate highest mole fraction and we know that lowering of vapor pressure is equal to mole fraction.
So, lowering of vapor pressure is observed more in case of urea than glucose, ${\rm{MgS}}{{\rm{O}}_{\rm{4}}}$ and ${\rm{BaC}}{{\rm{l}}_{\rm{2}}}$.
Hence, option A is correct.
Note: Relative lowering of vapor pressure depends on concentration of solute. In the question, molalities of solutes are given. So, the solute whose concentration is greater that is, urea has the highest lowering of vapor pressure.
Complete step-by-step answer:
First, we have to establish the relation between lowering of vapor pressure and mole fraction.
We know that the relation between mole fraction and vapour pressure of solvent and vapor pressure of solution is:
${p_1} = {x_1}{p_1}^0$ …… (1)
Where, ${p_1}$is the vapour pressure of solution, ${x_1}$is mole fraction of solvent and ${p_1}^0$ is vapour pressure of solvent.
The reduction in vapour pressure of solvent $\left( {\Delta {p_1}} \right)$ is given as:
$\Delta {p_1} = {p_1}^0 - {p_1}$
We have to put the value of ${p_1}$ from equation (1) in the below equation.
$\begin{array}{c}\Delta {p_1} = {p_1}^0 - {p_1}^0{x_1}\\ = {p_1}^0\left( {1 - {x_1}} \right)\end{array}$ …… (2)
We know that, ${x_1} + {x_2} = 1$, where ${x_1}$ is mole fraction of solvent and ${x_2}$ is mole fraction of solute.
$\begin{array}{c}{x_1} + {x_2} = 1\\{x_2} = 1 - {x_1}\end{array}$
Now, we have to put the value of $1 - {x_1}$ in equation (2).
$\Delta {p_1} = {x_2}{p_1}^0$ …… (3)
Now, we divide equation (3) by ${p_1}^0$.
$\begin{array}{c}\dfrac{{\Delta {p_1}}}{{{p_1}^0}} = \dfrac{{{x_2}{p_1}^0}}{{{p_1}^0}}\\\dfrac{{\Delta {p_1}}}{{{p_1}^0}} = {x_2}\\\dfrac{{{p_1}^0 - {p_1}}}{{{p_1}^0}} = \dfrac{{{n_2}}}{{{n_2} + {n_1}}}\end{array}$ ……. (4)
So, the relation establishes that lowering of vapor pressure is equal to mole fraction of solute. Mole fraction again related to moles of solute. So, we have to find the solution of highest molality to identify the solution possessing in which lowering of vapour pressure is highest.
In the given question, molality of urea is $0.2\;{\rm{m}}$, molality of glucose is 0.1 m, molality of ${\rm{MgS}}{{\rm{O}}_{\rm{4}}}$ is 0.1 m and molality of ${\rm{BaC}}{{\rm{l}}_{\rm{2}}}$ is 0.1 m.
The highest molality of urea indicates that moles of urea are highest among all the options.
Highest moles of solute indicate highest mole fraction and we know that lowering of vapor pressure is equal to mole fraction.
So, lowering of vapor pressure is observed more in case of urea than glucose, ${\rm{MgS}}{{\rm{O}}_{\rm{4}}}$ and ${\rm{BaC}}{{\rm{l}}_{\rm{2}}}$.
Hence, option A is correct.
Note: Relative lowering of vapor pressure depends on concentration of solute. In the question, molalities of solutes are given. So, the solute whose concentration is greater that is, urea has the highest lowering of vapor pressure.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)