Question

At 298 K, 0.714 moles of liquid A is dissolved in 5.555 moles of liquid B. The vapour pressure of the resultant solution is 475 torr. The vapour pressure of pure liquid A at the same temperature is 280.7 torr. What is the vapour pressure of pure liquid B in torr?

• A. 486
• B. 550
• C. 514
• D. 500

Hint:

Raoult's Law for a binary solution of volatile liquids: \( P_{total} = P_A^0 X_A + P_B^0 X_B \). Where \(X_A = \frac{n_A}{n_A+n_B}\) and \(X_B = \frac{n_B}{n_A+n_B}\) are mole fractions. \(P_A^0, P_B^0\) are vapour pressures of pure components. Substitute given values and solve for the unknown \(P_B^0\). Numbers like 0.714... might suggest fractions like 5/7 or related to 1/7 series. 5.555... = 50/9. Using these can make calculations exact if intended.

Answer 1

The Correct Option is D

According to Raoult's Law for a solution of two volatile liquids A and B, the total vapour pressure \( P_{total} \) of the solution is given by: \[ P_{total} = P_A^0 X_A + P_B^0 X_B \] where \( P_A^0 \) and \( P_B^0 \) are the vapour pressures of pure liquids A and B, respectively, and \( X_A \) and \( X_B \) are their mole fractions in the solution.
Given: Moles of A, \( n_A = 0.
714 \) mol.
Moles of B, \( n_B = 5.
555 \) mol.
Total moles \( n_{total} = n_A + n_B = 0.
714 + 5.
555 = 6.
269 \) mol.
Mole fraction of A, \( X_A = \frac{n_A}{n_{total}} = \frac{0.
714}{6.
269} \).
Mole fraction of B, \( X_B = \frac{n_B}{n_{total}} = \frac{5.
555}{6.
269} \).
Also, \( X_A + X_B = 1 \), so \( X_B = 1 - X_A \).
Let's calculate \(X_A\): \( X_A = \frac{0.
714}{6.
269} \).
Notice that \( 0.
714 \approx 0.
7 \).
\( 6.
269 \approx 6.
3 \).
\( 0.
7/6.
3 = 7/63 = 1/9 \).
Let's check if the numbers are related to common molar masses.
Molar mass of water (H\(_2\)O) is 18 g/mol.
\( 5.
555 \text{ moles} \approx 100\text{g}/18\text{g/mol} \).
So, B could be water if 100g was taken.
\( 0.
714 \text{ moles} \).
If it was acetone (58 g/mol), \(0.
714 \times 58 \approx 41.
4\)g.
Let's use the values given: \( X_A = \frac{0.
714}{6.
269} \).
Approximation \( 1/9 \approx 0.
1111 \).
\( 0.
714 / 6.
269 \approx 0.
11389 \).
If \(X_A \approx 1/9\), then \(X_B \approx 8/9\).
Let's test this: \( 0.
714 \times 9 = 6.
426 \).
So \( X_A \) is slightly less than \(6.
426 / (9 \times 6.
269) \).
This implies the numbers might be chosen for simpler fractions.
\( 0.
714 \approx 5/7 \times 1/ (5/7 + 50/9)\).
No.
If \(n_B/n_A = 5.
555/0.
714 \approx 7.
779 \).
So roughly \(n_B = 7.
8 n_A\).
Let's use \(X_A \approx 0.
114\) and \(X_B \approx 1 - 0.
114 = 0.
886\).
Given: \( P_{total} = 475 \) torr.
\( P_A^0 = 280.
7 \) torr.
We need to find \( P_B^0 \).
\[ 475 = (280.
7) X_A + P_B^0 X_B \] \[ 475 = (280.
7) \left(\frac{0.
714}{6.
269}\right) + P_B^0 \left(\frac{5.
555}{6.
269}\right) \] Multiply by 6.
269: \[ 475 \times 6.
269 = 280.
7 \times 0.
714 + P_B^0 \times 5.
555 \] \( 475 \times 6.
269 \approx 2977.
825 \) \( 280.
7 \times 0.
714 \approx 200.
39 \).
(More precisely 200.
3998) \[ 2977.
825 = 200.
3998 + 5.
555 P_B^0 \] \[ 5.
555 P_B^0 = 2977.
825 - 200.
3998 = 2777.
4252 \] \[ P_B^0 = \frac{2777.
4252}{5.
555} \] Approximate \( 2777 / 5.
55 \approx 2777 / (50/9) = 2777 \times 9 / 50 \approx 25000/50 = 500 \).
Let's check if \(P_B^0 = 500\) torr works with simpler fractions.
If \(X_A = 1/9\) and \(X_B = 8/9\): \( P_{total} = P_A^0 (1/9) + P_B^0 (8/9) \) \( 475 = (280.
7)/9 + P_B^0 (8/9) \) \( 475 \times 9 = 280.
7 + 8 P_B^0 \) \( 4275 = 280.
7 + 8 P_B^0 \) \( 8 P_B^0 = 4275 - 280.
7 = 3994.
3 \) \( P_B^0 = 3994.
3 / 8 \approx 499.
2875 \).
This is very close to 500.
The numbers \(0.
714\) and \(5.
555\) might be specifically chosen.
Note \( 5.
555.
.
.
= 50/9 \).
And \( 0.
7142857.
.
.
= 5/7 \).
If \( n_A = 5/7 \) and \( n_B = 50/9 \).
\( n_{total} = \frac{5}{7} + \frac{50}{9} = \frac{45+350}{63} = \frac{395}{63} \).
\( X_A = \frac{5/7}{395/63} = \frac{5}{7} \times \frac{63}{395} = \frac{5 \times 9}{395} = \frac{45}{395} = \frac{9}{79} \).
\( X_B = \frac{50/9}{395/63} = \frac{50}{9} \times \frac{63}{395} = \frac{50 \times 7}{395} = \frac{350}{395} = \frac{70}{79} \).
Using these exact fractions: \( 475 = 280.
7 \times \frac{9}{79} + P_B^0 \times \frac{70}{79} \) \( 475 \times 79 = 280.
7 \times 9 + 70 P_B^0 \) \( 37525 = 2526.
3 + 70 P_B^0 \) \( 70 P_B^0 = 37525 - 2526.
3 = 34998.
7 \) \( P_B^0 = \frac{34998.
7}{70} \approx 499.
98 \).
This is extremely close to 500 torr.
The initial values were likely rounded versions of these fractions.
Therefore, \( P_B^0 = 500 \) torr.
This matches option (4).