Question

The vapour pressure of a saturated solution of sparingly soluble salt (\[XC{{l}_{3}}\]) was 17.20 mm Hg at $27^oC$ . If the vapour pressure of pure \[{{H}_{2}}O~\] is 17.25 mm Hg at 300 K, what is the solubility of sparingly soluble salt \[XC{{l}_{3}}\] in mole per Litre?
A. \[4.04\times {{10}^{-2}}\]
B. \[8.08\times {{10}^{-2}}\]
C. \[2.02\times {{10}^{-2}}\]
D. \[4.04\times {{10}^{-3}}\]

Answer 1

Hint: To solve this question we should know that solubility is a property referring to the ability for a given substance, the solute, to dissolve in a solvent. As the salt is sparingly soluble, molarity will be equal to molality.

Step by step answer:
We should know that solubility is defined as the maximum amount of a substance that will dissolve in a given amount of solvent at a specified temperature. Solubility is a characteristic property of a specific solute–solvent combination, and different substances have greatly differing solubility.
To solve this question, we will do following steps:
Let solubility of \[XC{{l}_{3}}\] = S mole per litre.
We should then find “n” which is the number of particles in a formula that is three chlorine atoms and one X atom. So, value of “n” will be:
$n=3+1=4$
We are assuming that it is a 100% ionized solution.
$\alpha =1$
$\begin{align}
  & i=1+(n-1)\alpha \\
 & \,\,\,=1+(4-1)1 \\
 & i\,=4 \\
\end{align}$
As it is given in the question, solubility is very less so, Molarity will become equal to molality.
\[~4.04\times {{10}^{-3}}\]\[\dfrac{P{}^\circ -{{P}_{solution}}}{P{}^\circ }=i\times \dfrac{moles\,of\,solute}{moles\,of\,solvent}\]
$\begin{align}
  & P{}^\circ =\,\,vapour\,pressure\,of\,{{H}_{2}}O=\,17.25\,mmHg \\
 & {{P}_{solution}}=\,vapour\,pressure\,\,of\,solution=17.20mmHg \\
\end{align}$
Moles of solute= S
Moles of solvent= \[\dfrac{mass\,of\,water}{molar\,mass}=\dfrac{1000}{18}\]
\[\begin{align}
  & \dfrac{17.25-17.20}{17.25}=4\times \dfrac{S}{\dfrac{1000}{18}} \\
 & \dfrac{0.05}{17.25}=\dfrac{4}{55.55}\times S \\
 & S=\dfrac{0.05\times 55.55}{4\times 17.25} \\
 & S=4.04\times {{10}^{-2}}\dfrac{mole}{litre} \\
\end{align}\]
So, from the above calculations we should know that our correct answer is option A.

Note: We should know that vapour pressure of a liquid is the equilibrium pressure of a vapour above its liquid (or solid); that is, the pressure of the vapour resulting from evaporation of a liquid (or solid) above a sample of the liquid (or solid) in a closed container. We should note that when a solid or a liquid evaporates to a gas in a closed container, the molecules cannot escape. Some of the gas molecules will eventually strike the condensed phase and condense back into it. When the rate of condensation of the gas becomes equal to the rate of evaporation of the liquid or solid, the amount of gas, liquid and/or solid no longer changes. The gas in the container is in equilibrium with the liquid or solid.