The mass of a non-volatile, non-electrolyte solute (molar mass =   50   g   mol - 1 ) needed…

Molar mass of octane = 114 g/mol.

From the lowering of vapour pressure, we have,

$\frac{∆\mathrm{P}}{\mathrm{P}}=\frac{\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2}}{\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2}+\frac{{\mathrm{W}}_1}{{\mathrm{M}}_1}}$

Where ${\mathrm{W}}_2\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }{\mathrm{M}}_2$ are mass and molar mass of solute
and ${\mathrm{W}}_1\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }{\mathrm{M}}_1$ are mass and molar mass of octane.

$\frac{75}{100}=\frac{\frac{{\mathrm{W}}_2}{50\mathrm{ }\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}}}{\frac{{\mathrm{W}}_2}{50\mathrm{ }\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}}+\frac{114\mathrm{ }\mathrm{g}}{114\mathrm{ }\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }}}$

$0.75=\frac{\frac{{\mathrm{W}}_2}{50}}{\frac{{\mathrm{W}}_2}{50}+1}$

$\frac{{\mathrm{W}}_2}{50}+1=\frac{{\mathrm{W}}_2}{50\times 0.75}$

${\mathrm{W}}_2=150\mathrm{g}$