An aqueous solution containing 64% by weight of volatile liquid ‘A’ (molecular mass 128) has pressure of 145mm, then vapour pressure of ‘A’ is (V.P. of H₂O is 155 mm)

We are tasked with finding the vapor pressure of liquid A (denoted as P_A⁰) in an aqueous solution. To do this, we will apply Raoult's Law, which connects the vapor pressures of the components in a solution with their mole fractions.

1. Given Data:

• Weight percentage of liquid A = 64%
• Vapor pressure of the solution (P_t) = 145 mm Hg
• Vapor pressure of water (P_B⁰) = 155 mm Hg
• Molecular weight of liquid A = 128 g/mol
• Molecular weight of water = 18 g/mol

2. Calculate the Weight of Components:

Assume the total weight of the solution is 100 g.

• Weight of liquid A = 64 g
• Weight of water = 100 g - 64 g = 36 g

3. Calculate the Moles of Each Component:

• Moles of liquid A = Weight of A / Molecular weight of A = 64 g / 128 g/mol = 0.5 mol
• Moles of water = Weight of water / Molecular weight of water = 36 g / 18 g/mol = 2 mol

4. Calculate the Total Moles in the Solution:

Total moles = Moles of liquid A + Moles of water = 0.5 mol + 2 mol = 2.5 mol

5. Calculate the Mole Fractions:

• Mole fraction of liquid A (X_A) = Moles of A / Total moles = 0.5 mol / 2.5 mol = 0.2
• Mole fraction of water (X_B) = 1 - X_A = 1 - 0.2 = 0.8

6. Apply Raoult's Law:

According to Raoult's Law, the total vapor pressure of the solution is given by the equation:

P_t = P_A⁰ × X_A + P_B⁰ × X_B

Substitute the known values into the equation:

145 mm Hg = P_A⁰ × 0.2 + 155 mm Hg × 0.8

7. Calculate the Vapor Pressure of Liquid A (P_A⁰):

First, calculate 155 mm Hg × 0.8:

155 mm Hg × 0.8 = 124 mm Hg

Now, substitute this value back into the equation:

145 mm Hg = P_A⁰ × 0.2 + 124 mm Hg

Rearranging the equation gives:

P_A⁰ × 0.2 = 145 mm Hg - 124 mm Hg = 21 mm Hg

Finally, solve for P_A⁰:

P_A⁰ = 21 mm Hg / 0.2 = 105 mm Hg

Final Answer:

The vapor pressure of liquid A is 105 mm Hg.