Question

In ideal solution of non volatile solute in solvent A in 2:5 molar ratio has vp 250 mm. If a another solution in ratio 3:4 prepared then vp above this solution

Answer 1

200

Answer 2

Let $P_A^0$ be the vapor pressure of pure solvent A.

For the first solution, the molar ratio of non-volatile solute to solvent A is 2:5. The mole fraction of solvent A is $x_{A,1} = \frac{\text{moles of A}}{\text{total moles}} = \frac{5}{2+5} = \frac{5}{7}$. The vapor pressure of the first solution is given as $P_1 = 250$ mm.

According to Raoult's Law for an ideal solution with a non-volatile solute, the vapor pressure of the solution is proportional to the mole fraction of the solvent: $P_1 = x_{A,1} \cdot P_A^0$ $250 = \frac{5}{7} \cdot P_A^0$

Solving for $P_A^0$: $P_A^0 = \frac{250 \times 7}{5} = 50 \times 7 = 350$ mm.

Now consider the second solution, where the molar ratio of solute to solvent A is 3:4. The mole fraction of solvent A in the second solution is $x_{A,2} = \frac{\text{moles of A}}{\text{total moles}} = \frac{4}{3+4} = \frac{4}{7}$. The vapor pressure of the second solution, $P_2$, can be calculated using Raoult's Law: $P_2 = x_{A,2} \cdot P_A^0$

Using the value of $P_A^0 = 350$ mm: $P_2 = \frac{4}{7} \times 350$ $P_2 = 4 \times \left(\frac{350}{7}\right)$ $P_2 = 4 \times 50$ $P_2 = 200$ mm.

The vapor pressure above the second solution is 200 mm.