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Lowering in vapour pressure is highest for:
A. 0.2 m urea
B. 0.1 m glucose
C. 0.1 m ${\rm{MgS}}{{\rm{O}}_{\rm{4}}}$
D. 0.1 m ${\rm{BaC}}{{\rm{l}}_{\rm{2}}}$

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Last updated date: 23rd May 2024
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Answer
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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.