$\begin{array}{1 1}(a)\;36.1kPa\\(b)\;96.2kPa\\(c)\;144.5kP\\(d)\;72.0kPa\end{array}$

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Answer: 72 kPa

According to Daltonâ€™s law of partial pressures, the total pressure ( total p ) over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as: $p_T = p_A + p_B$

In our case $p_T = p_{\text{Octane}} + p_{\text{Heptane}}$; For simplicity we'll refer to Octane and Heptane as A and B.

Raoult's law states that for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.

Combining the two laws, we get: $p_T = p_A^0\;x_A + p_B^0\;x_B$

Given that at 373K,the vapour pressures of the two liquid components (heptane and octane) are 105kPa and 45kPa respectively, $\Rightarrow p_T = 105 x_A + 45 x_B$

$\Rightarrow x_A = \large\frac{n_A}{n_A + n_B} $$ = \large\frac{25/100}{25/100+35/114} $$ = 0.449$

$\Rightarrow x_B = \large\frac{n_B}{n_A + n_B} $$ = \large\frac{35/114}{25/100+35/114} $$ = 0.551$

$\Rightarrow p_T = 105 \times 0.449 +45 \times 0.551 = 47.25 + 24.75 \approx 72\;kPa$

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