Two liquids A and B form an ideal solution. At 300K the vapour pressure of a solution of 1 mole of A and x moles of B is 550mm. If the vapour pressures of pure A and B are 400mm and 600mm respectively, then x is:

• A) 1
• B) 2
• C) 3
• D) 4

To find the value of x in the given problem, we can use Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the vapor pressures of its components, each multiplied by its mole fraction in the solution.

Given Data

• Vapor pressure of pure A (P_A) = 400 mm
• Vapor pressure of pure B (P_B) = 600 mm
• Vapor pressure of the solution (P_solution) = 550 mm
• Moles of A = 1
• Moles of B = x

Calculating Mole Fractions

The total number of moles in the solution is:

Total moles = 1 + x

The mole fraction of A (X_A) is:

X_A = 1 / (1 + x)

The mole fraction of B (X_B) is:

X_B = x / (1 + x)

Applying Raoult's Law

According to Raoult's Law:

P_solution = X_A * P_A + X_B * P_B

Substituting the values:

550 = (1 / (1 + x)) * 400 + (x / (1 + x)) * 600

Simplifying the Equation

Multiply through by (1 + x) to eliminate the denominator:

550(1 + x) = 400 + 600x

Expanding gives:

550 + 550x = 400 + 600x

Rearranging terms leads to:

150 = 50x

Thus, x = 3.

Final Answer

The value of x is 3, so the correct option is C) 3.