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An aqueous solution of $4\%$ non-volatile solute exerts a pressure of $1.004bar$ at the normal boiling point of the solvent. What is the molar mass of the solute? $\left( A \right)41.35gmo{l^{ - 1}}$$\left( B \right)60.21gmo{l^{ - 1}}$$\left( C \right)84.42gmo{l^{ - 1}}$$\left( D \right)56.91gmo{l^{ - 1}}$

Last updated date: 19th Jun 2024
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Hint: A solution is made up of solvent and solute. non-volatile solute means which does not evaporate . It has high boiling points and less vapour pressure . we can also say that substances which have boiling point ${100^ \circ }C$ or more are termed as volatile rest all are non-volatile .

Vapour pressure of water at normal boiling point given is $1.013bar$
Vapour pressure given for solution in normal boiling point is $1.004bar$
Given that solute $4\%$ means $4g$
Mass of solvent will be $100 - 4 = 96g$
Molar mass of water is $18gmo{l^{ - 1}}$
Now we have to apply Raoult’s law
$\Rightarrow$ $\dfrac{{{p_1}^0 - {p_1}}}{{{p_1}^0}} = \dfrac{{{w_2} \times {M_1}}}{{{M_2} \times {w_1}}}$
Which states that vapour pressure of a solution of non-volatile solute should be equal to the vapour pressure of the pure solvent at that temperature which is multiplied with mole fraction.
Putting all the values
$\Rightarrow$ $\dfrac{{1.013 - 1.004}}{{1.013}} = \dfrac{{2 \times 18}}{{{M_2} \times 96}}$
${M_2} = 84.42gmo{l^{ - 1}}$
Hence the molar mass of the solute is $84.42gmo{l^{ - 1}}$

So the correct answer is Option B.