Question

A saturated solution of sparingly soluble salt \[MC{l_2}\]has a vapour pressure of \[31.78{\text{mm}}\] of \[Hg\] at \[{30^\circ }C\]. Pure water exerts a pressure of \[31.28{\text{mm}}\] of \[Hg\] at \[{30^\circ }C\]. The \[{K_{sp}}\] of \[MC{l_2}\] is \[X \times {10^{ - 5}}\]\[mo{l^3}\]\[litr{e^{ - 3}}\], value of \[X\] will be.
Assume molarity equal to molarity and \[MC{l_2}\] is \[100\% \] dissociated in solution.

Answer 1

Hint:
Sparingly soluble materials are those which have lowered solubility. Usually, materials are treated as sparingly soluble if \[1gm\] of material requires \[30\] to \[100{\text{ml}}\] of solvent to dissolve.

Complete step by step answer:
One of the colligative property is ‘Relative lowering of vapour pressure’ \[\left( {{\text{R}}{\text{.L}}{\text{.V}}{\text{.P}}} \right)\]
\[{\text{R}}{\text{.L}}{\text{.V}}{\text{.P}} = \dfrac{{P_A^\circ - {P_S}}}{{P_A^\circ }} = {X_B}\]
The relative lowering of vapour pressure of solvent is equal to the mole fraction of the solute.
According to the question,
Assume \[{\text{1 litre}}\] of solutions. It contains \[55.56\] moles of water
\[\dfrac{{{P^\circ } - P}}{{{P^\circ }}} = \dfrac{{nMC{l_2}}}{{n{H_2}O}}\]
\[\dfrac{{31.82 - 31.72}}{{31.82}} = \dfrac{{nMC{l_2}}}{{55.56}}\]
\[nMC{l_2} = 6.98 \times {10^{ - 2}}\] moles
Since, these moles are present in \[{\text{1 litre}}\] they account for the molarity of solution.
But we know that \[MC{l_2}\]is \[100\% \]ionized in solution, it gives \[3\] ions per molecule. Hence, they will one third of \[6.98 \times {10^{ - 2}}\] which comes out to be \[0.0349M\].
\[MC{l_2}\left( s \right) = {M^{2 + }}\left( {aq.} \right) \times 2C{l^ - }\left( {aq.} \right)\]
The expression for the solubility product is:
\[{K_{sp}} = \left[ {{M^{2 + }}} \right]{\left[ {C{l^ - }} \right]^2} = \left( S \right){\left( {2S} \right)^2}\]
\[ = 4{S^3} = 4{\left( {0.349} \right)^3}\]
\[ = 1.7 \times {10^{ - 4}}\]
\[{K_{sp}} = 1.7 \times {10^{ - 5}}\]
Comparing it with \[{K_{sp}} = X \times {10^{ - 5}}\]
\[X = 17\]
So, \[17\] is the answer to this question.

Additional information:
Since the solute molecules are non-volatile, the vapour above the solution consists of only solvent (pure liquid) molecules. After adding the solute, the vapour pressure of the solution is found to be lower than that of pure liquid at room temperature.
This is known as “Relative lowering of vapour pressure”.
The decrease in vapour pressure of the liquid is proportional to the concentration of particles of solute, i.e. to the number of dissolved molecules or ions per unit volume.
This is a colligative property which is used to determine the molar mass of a solute, the following expression is used:
$\dfrac{{\Delta P}}{{P{1^o}}} = \dfrac{{{W_2}{M_1}}}{{{W_1}{M_2}}}$
${W_1}{\text{ and }}{W_2}$ are Moller whereas ${M_1}{\text{ and }}{M_2}$ are masses whereas are molar masses of solvent and solute respectively.

Note: Colligative properties are those properties of solutions that depend upon the ratio of number solvent molecules in a solution.It does not depend on the nature of the chemical species present.