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.