From the lowering of vapour pressure we have,

$\frac{{\Delta P}}{P} = \frac{{\frac{{{W_2}}}{{{M_2}}}}}{{\frac{{{W_2}}}{{{M_2}}} + \frac{{{W_1}}}{{{M_1}}}}}$

$\frac{{75}}{{100}} = \frac{{\frac{{{W_2}}}{{50\,g/mol}}}}{{\frac{{{W_2}}}{{50\,g/mol}} + \frac{{114\,g}}{{114\,g/mol}}}}$

$0.75 = \frac{{\frac{{{W_2}}}{{50\,}}}}{{\frac{{{W_2}}}{{50\,}} + 1}}$

$\frac{{{W_2}}}{{50}} + 1\, = \,\frac{{{W_2}}}{{50\, \times \,0.75}}$

${W_2}\, = \,150\,g$

Note : $W_2$ vand $M_2$ are mass and molar mass of solute and $W_1$ and $M_1$ are mass and molar mass of octane.