Question

The lowering in vapour pressure is maximum for:
(A)- $\text{0}\text{.1}\,\text{M}\,\text{urea}$
(B)- $\text{0}\text{.1}\,\text{M}\,\text{NaCl}$
(C)- $\text{0}\text{.1}\,\text{M}\,\text{MgC}{{\text{l}}_{2}}$
(D)- $\text{0}\text{.1}\,\text{M}\,{{K}_{4}}\left[ Fe{{(CN)}_{6}} \right]$

Answer 1

Hint: The lowering in vapour pressure is caused by the addition of solute particles into the solvent. As the number of solute particles increases, the vapour pressure of the solution decreases.

Complete step by step answer:
The lowering of vapour pressure is a colligative property which depends on the amount of solute in the solution. That is, the number of solute particles present in the solution by dissociation or association. In the given solutions, having equal concentration of 0.1 M, the solute component is ionic. It produces ions when dissolved in solvent. Thus, increasing the number of particles in the solution. This is measured using the Van’t Hoff factor $(i)$, which is the ratio of the number of solute ions produced to its actual concentration (in terms of mass).
$i=\dfrac{\text{apparent number of ions produced}}{\text{number of moles of solute dissolved}}$. So, in case of ionic compounds, the Van’t Hoff factor is equal to the number of ions produced on dissociation of one formula unit of the substance. Then, the Van’t Hoff factor are as follows:
- Urea does not dissociate, so $i=1$
- The Sodium chloride molecule dissociates into $NaCl\to N{{a}^{+}}+C{{l}^{-}}$, so two moles of ions are produced. Then, $i=2$
- Magnesium chloride molecule dissociates into $MgC{{l}_{2}}\to M{{g}^{2+}}+2C{{l}^{-}}$. Here, one mole of magnesium ion and two moles of chloride ions are produced. So, $i=3$.
- Potassium ferrocyanide molecule dissociates into ${{K}_{4}}\left[ Fe{{(CN)}_{6}} \right]\to 4{{K}^{+}}+{{\left[ Fe{{(CN)}_{6}} \right]}^{4-}}$. Here four moles of ${{K}^{+}}$ ions and one mole of ${{\left[ Fe{{(CN)}_{6}} \right]}^{4-}}$ is produced. So, $i=5$.

Thus, the potassium ferrocyanide molecule produces the maximum number of solute ions in the solution. So, as the number of solute particles increases in the solution, the vapour pressure decreases compared to the pure solvent.

Therefore, the lowering in vapour pressure is maximum for option (D)- $\text{0}\text{.1}\,\text{M}\,{{K}_{4}}\left[ Fe{{(CN)}_{6}} \right]$.

Note: The vapour pressure is based on Raoult's law, where the partial vapor pressure of each component in the solution is proportional to its mole fraction in the mixture. This mole fraction tells us about the number of particles present.