Question

The vapour pressure of ethyl alcohol at ${{25}^{\text{ 0}}}\text{C}$is $\text{59}\text{.2 torr}$. The vapour pressure of a solution of non-volatile solute urea, $\text{51}\text{.3 Torr}$. What is the molarity of the solution?

Answer 1

This question is related to the relative lowering of vapour pressure of solvents on the addition of non-volatile solutes to them to form a solution. This property is called the Colligative Property which is dependent upon the number of the solute molecules present in the medium and not on the nature of the solute.

Formula Used: As per the Raoult’s Law, the relative lowering of vapour pressure can be written as,
$\dfrac{\text{P - }{{\text{P}}_{\text{s}}}}{\text{P}}\text{=}\dfrac{\text{n}}{\text{n+N}}$; where ${{\text{P}}_{\text{s}}}$is the vapour pressure of the solution, $\text{P}$ is the vapour pressure of the pure solvent, n is the number of moles of the solute present in the solution and N is the number of moles of the solvent in the solution.

Complete step by step answer:
Given that, $\text{P}$ is the vapour pressure of the pure solvent =$\text{59}\text{.2 torr}$
${{\text{P}}_{\text{s}}}$ is the vapour pressure of the solution = $\text{51}\text{.3 Torr}$
The relative lowering of vapour pressure = $\dfrac{\text{P - }{{\text{P}}_{\text{s}}}}{\text{P}}=\dfrac{\text{59}\text{.2 -51}\text{.3 }}{\text{59}\text{.2 }}=\dfrac{7.9}{59.2}=0.133$
The molecular weight of urea= $\left[ \left( 14\times 2 \right)+\left( 1\times 4 \right)+12+16 \right]=60$gram
The molecular weight of ethyl alcohol = $\left[ \left( 12\times 2 \right)+\left( 1\times 6 \right)+16 \right]=46$gram
The molality of a solution = number of moles of the solute per kilogram of the solvent.
$\dfrac{\text{n}}{\text{n + N}}\text{ = }{{\text{x}}_{\text{2}}}$ Where$\text{ }{{\text{x}}_{\text{2}}}$ is the mole fraction of the solute and
$\dfrac{\text{N}}{\text{n + N}}\text{ = }{{\text{x}}_{\text{1}}}$ Where ${{\text{x}}_{\text{1}}}$ is the mole fraction of the solvent.
Now the molality of the solute can also be written as, $\text{m =}\dfrac{{{\text{x}}_{2}}}{\text{1-}{{\text{x}}_{2}}}\text{ }\times\text{ }\dfrac{\text{1000}}{{{\text{M}}_{1}}}$
Putting the values in the above equation, we get,
$\text{m =}\dfrac{0.133}{\text{1-0}\text{.133}}\text{ }\times\text{ }\dfrac{\text{1000}}{46}$=$3.33$

Therefore, the molality of the solution is equal to$3.33$.
Note:
The molality of a solution is a unit to express the solution concentration and is expressed as the number of moles of the solute per kilogram of the solvent. Unlike other units to express the solution's concentration, the molality of the solution is dependent upon the temperature of the solution and the volume of the solution is dependent upon the temperature of the solution.