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Step-by-Step Solution
Step 1: Identify the Known Information
We have a solution of n-hexane and n-heptane at 300 K. Initially, the solution contains:
1 mole of n-hexane
3 moles of n-heptane
The total vapour pressure of this solution is given as 550 mmHg. When 1 more mole of n-heptane is added (making it 4 moles of n-heptane in total), the new vapour pressure becomes 560 mmHg. We are asked to find the vapour pressure of pure n-heptane, denoted as P_2^0 .
Step 2: Apply Raoult’s Law
According to Raoult’s Law, the total vapour pressure P_{\text{total}} of a two-component solution (components 1 and 2) is:
P_{\text{total}} = P_1^0 \, X_1 + P_2^0 \, X_2
where
P_1^0 is the vapour pressure of pure n-hexane.
P_2^0 is the vapour pressure of pure n-heptane.
X_1 and X_2 are the mole fractions of n-hexane and n-heptane in the solution, respectively.
Step 3: Write the Expressions for the Initial Solution
Initially, the solution has 1 mole of n-hexane and 3 moles of n-heptane, so the total number of moles is 1 + 3 = 4 .
Their mole fractions are:
X_1 = \dfrac{1}{4}, \quad X_2 = \dfrac{3}{4}
Thus, the total vapour pressure of 550 mmHg can be written as:
550 = P_1^0 \times \dfrac{1}{4} + P_2^0 \times \dfrac{3}{4}
Multiply through by 4 to simplify:
2200 = P_1^0 + 3 \, P_2^0\quad \dots (1)
Step 4: Write the Expression after Adding One More Mole of n-Heptane
When 1 more mole of n-heptane is added, the new composition is:
1 mole of n-hexane
4 moles of n-heptane
The total number of moles is now 1 + 4 = 5 .
The new mole fractions are:
X_1 = \dfrac{1}{5}, \quad X_2 = \dfrac{4}{5}
Given that the new vapour pressure is 560 mmHg:
560 = P_1^0 \times \dfrac{1}{5} + P_2^0 \times \dfrac{4}{5}
Multiply through by 5 to simplify:
2800 = P_1^0 + 4 \, P_2^0 \quad \dots (2)
Step 5: Solve the System of Equations
We have two linear equations:
(1)\quad P_1^0 + 3 P_2^0 = 2200
(2)\quad P_1^0 + 4 P_2^0 = 2800
Subtract equation (1) from equation (2) to find P_2^0 :
(P_1^0 + 4 P_2^0) - (P_1^0 + 3 P_2^0) = 2800 - 2200
This simplifies to:
P_2^0 = 600 \text{ mmHg}
Substitute P_2^0 = 600 back into either equation (1) or (2) to find P_1^0 if needed:
P_1^0 + 3(600) = 2200 \implies P_1^0 = 2200 - 1800 = 400 \text{ mmHg}
Step 6: Conclude the Vapour Pressure of Pure n-Heptane
The vapour pressure of pure n-heptane P_2^0 is 600 mmHg.
