The vapour pressure of two pure liquids A and B are 50 and 40 torr respectively. If 8 moles of A is mixed with x moles of B, then vapour pressure of solution obtained is 48 torr. What is the value of x.
Question
The vapour pressure of two pure liquids A and B are 50 and 40 torr respectively. If 8 moles of A is mixed with x moles of B, then vapour pressure of solution obtained is 48 torr. What is the value of x.
Solution
This problem can be solved using Raoult's Law, which states that the partial pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.
Given:
- Vapor pressure of pure A (P_A) = 50 torr
- Vapor pressure of pure B (P_B) = 40 torr
- Total vapor pressure of the solution (P_total) = 48 torr
- Moles of A (n_A) = 8 moles
- Moles of B (n_B) = x moles
According to Raoult's Law, the total vapor pressure of the solution is the sum of the partial pressures of A and B:
P_total = P_A * (n_A / (n_A + n_B)) + P_B * (n_B / (n_A + n_B))
Substituting the given values:
48 = 50 * (8 / (8 + x)) + 40 * (x / (8 + x))
Solving this equation for x will give the number of moles of B.
First, simplify the equation:
48 = 400 / (8 + x) + 40x / (8 + x)
Multiply through by (8 + x) to clear the denominator:
48 * (8 + x) = 400 + 40x
Distribute on the left side:
384 + 48x = 400 + 40x
Subtract 40x from both sides:
8x = 16
Finally, solve for x:
x = 2
So, 2 moles of B are needed.
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