When a certain amount of solid A is dissolved in 100 g of water at 25°C to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is 23.76 mmHg. The number of moles of solute A added is________. (Nearest Integer)

Complete Solution:

Given Data:

• Vapour pressure of pure water: P_water = 23.76 mmHg
• Vapour pressure of solution: P_solution = 11.88 mmHg
• Mass of water: 100 g
• Molar mass of water: 18 g/mol

The mole fraction of water (χ_water) is given by:

χ_water = P_solution / P_water

Substitute the values:

χ_water = 11.88 / 23.76 = 0.5

The mole fraction of water is:

χ_water = n_water / (n_water + n_A).

Rearranging to solve for n_A:

n_A = n_water (1 - χ_water) / χ_water

n_A = n_water × (1 / 2)

Moles of water:

n_water = mass of water / molar mass of water

n_water = 100 / 18 = 5.56 mol

Substitute into the formula:

n_A = 5.56 / 2 = 2.78 mol

Final Answer:

The number of moles of solute A added is 2.78 mol