Question

On mixing, heptane and octane form an ideal solution. At $373$ K, the vapour pressure of the two liquid components (heptane and octane) are $105$ kPa and $45$ kPa respectively. Vapour pressure of the solution obtained by mixing $25$ g of heptane and $35$ g of octane will be : (molar mass of heptane = $100$ g $mo{l}^{-2}$ and of octane = $114$ g $mo{l}^{-1}$)

• A. $96.2kPa$
• B. $144.5kPa$
• C. $72kPa$
• D. $36.1kPa$

Answer 1

Answer: (C)

$72kPa$

Heptane A, $n_A=\frac{25}{100}=0.25mol$

Octane B, $n_B=\frac{35}{114}=0.31mol$

$x_A=\frac{n_A}{n_A+n_B}=\frac{0.25}{0.25+0.31}=0.45$

$x_B=0.55$

Now,

$p_T=p_A^0{x}_A+p_B^0{x}_B$

$p_T=105\times 0.45+45\times 0.55$

$p_T=72kPa$