Question

100 g of liquid A \((molar\, mass\, 140\, g\, mol^{–1})\) was dissolved in 1000 g of liquid B \((molar\, mass\, 180\, g\, mol^{–1})\). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Answer 1

The correct answer is: 280.7 torr
Number of moles of liquid A, \(n_A=\frac{100}{140}mol\)
=0.714mol
Number of moles of liquid B, \(n_B=\frac{1000}{180}mol\)
=5.556mol
Then, mole fraction of A, \(x_A=\frac{n_A}{n_A+n_B}\)
\(=\frac{0.714}{0.714+5.556}\)
=0.114
And, mole fraction of B, \(x_B = 1 - 0.114\)
= 0.886 
Vapour pressure of pure liquid B, \(p^o_B = 500 torr \)
Therefore, vapour pressure of liquid B in the solution, 
\(p_B=p^o_Bx_B\)
\(= 500 × 0.886 \)
= 443 torr 
Total vapour pressure of the solution, ptotal = 475 torr 
∴Vapour pressure of liquid A in the solution, 
\(p_A = p_{total} - p_B\) 
= 475 - 443 
= 32 torr 
Now,
\(p_A=p^o_Ax_A\)
\(⇒p^o_A=\frac{p_A}{x_A}\)
\(=\frac{32}{0.114}\)
= 280.7 torr 
Hence, the vapour pressure of pure liquid A is 280.7 torr