Question

At 100$^{\circ }C$ the vapour pressure of a solution of 6.5g of a solute in 100g water is 732 mm. If K=0.52, the boiling point of this solution will be
A. 100$^{\circ }C$
B. 102$^{\circ }C$
C. 103$^{\circ }C$
D. 101$^{\circ }C$

Answer 1

Hint: Raolt’s law is used to solve this question as it links the change in the vapour pressure of a solution with the mole fraction of the solute. This makes it possible to find the molecular weight of the solute and once we find the molecular weight, we can easily find the change in the temperature of the solute which will lead us to the boiling point of the solute.

Complete step by step solution:
-It was observed that the vapour pressure of a solution was always less than the vapour pressure of a pure solvent. This was due to the addition of the solute particles in a solvent to form a solution.
- The addition of solute decreases the surface area of solvent and so evaporation decreases because the evaporation is a surface process and so lesser solvent evaporates resulting in lesser vapour pressure.
-So the vapour pressure depends on the amount of solute as well as the solvent. We can thus say that the total vapour pressure can be expressed by an equation ${{P}_{S}}={{P}_{A}}^{\circ }{{x}_{A}}+{{P}_{B}}^{\circ }{{x}_{B}}$ where x is the mole fraction and P is the vapour pressure of the solute and the solvent. A represents solvent and B represents solute.
-Also, for a non-volatile solute, we can say that ${{P}_{B}}^{\circ }=0\text{ and }{{\text{x}}_{A}}\text{+}{{\text{x}}_{B}}=1$ always. So we can write the above equation in terms of solvent alone as
\[\begin{align}
& {{P}_{S}}={{P}_{A}}^{\circ }{{x}_{A}} \\
& \Rightarrow \dfrac{1}{{{x}_{A}}}=\dfrac{{{P}_{A}}^{\circ }}{{{P}_{S}}} \\
& \Rightarrow \dfrac{{{P}_{A}}^{\circ }}{{{P}_{S}}}=\dfrac{{{n}_{A}}+{{n}_{B}}}{{{n}_{A}}}\text{ as }{{\text{x}}_{A}}=\dfrac{{{n}_{A}}}{{{n}_{A}}+{{n}_{B}}} \\
& \Rightarrow \dfrac{{{P}_{A}}^{\circ }-{{P}_{S}}}{{{P}_{S}}}=\dfrac{{{n}_{B}}}{{{n}_{A}}} \\
& \Rightarrow \dfrac{{{P}_{A}}^{\circ }-{{P}_{S}}}{{{P}_{S}}}=\dfrac{w}{m}.\dfrac{M}{W}\text{ as mole =}\dfrac{wt.\text{ in gram}}{molecular\text{ wt}\text{.}} \\
\end{align}\]
-This is the equation of the Raolt’s Law where W= wt. of solvent
w = wt. of solute
M = molecular mass of solvent
M = molecular mass of solute
${{P}_{S}}$ =vapour pressure of solution
${{P}_{A}}^{\circ }$ =vapour pressure of pure solvent
-We know that the vapour pressure of pure water is 760mm and using the formula we can write
\[\begin{align}
& \dfrac{760-732}{760}=\dfrac{6.5x18}{Mx100} \\
& \Rightarrow \text{M=}\dfrac{760x6.5x18}{100x28} \\
& \Rightarrow M=31.75gmo{{l}^{-1}} \\
\end{align}\]
-Now to find the rise in the temperature, we use the formula
$\begin{align}
& \Delta {{T}_{B}}={{K}_{B}}xm \\
& \Rightarrow \Delta {{T}_{B}}=0.52x\dfrac{6.5x1000}{31.75x100}={{1.06}^{\circ }}C \\
\end{align}$
We know the boiling point of water is $100$$^{\circ }C$ and we calculated the increment in the boiling point. Thus the boiling point of the solution will be $100+1.06$ $=101.06$$^{\circ }C$

Therefore the correct option is D.

Note: Always take care of the units of the quantities of temperature and pressure. Also keep in mind the formula of mole is calculated for the unit of mass as grams. If the vapour pressure of solution is taken in a particular unit, we need to take the vapour pressure of pure solvent in that same unit itself. For the above question, we took vapour pressure of water as 760mm as the pressure of solution was also given in mm