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

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

According to Dalton’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$
$\Rightarrow p_T = 105 \times 0.449 +45 \times 0.551 = 47.25 + 24.75 \approx 72\;kPa$
edited Jul 16, 2014