The mass of a non-volatile, non-electrolyte solute (molar mass =...

Molar mass of octane = 114 g/mol.

From the lowering of vapour pressure, we have,

$\dfrac{∆ P}{P} = \dfrac{\dfrac{{W}_{2}}{{M}_{2}}}{\dfrac{{W}_{2}}{{M}_{2}} + \dfrac{{W}_{1}}{{M}_{1}}}$

Where ${W}_{2}and{M}_{2}$ are mass and molar mass of solute
and ${W}_{1}and{M}_{1}$ are mass and molar mass of octane.

$\dfrac{75}{100} = \dfrac{\dfrac{{W}_{2}}{50g / mol}}{\dfrac{{W}_{2}}{50g / mol} + \dfrac{114g}{114g / mol}}$

$0.75 = \dfrac{\dfrac{{W}_{2}}{50}}{\dfrac{{W}_{2}}{50} + 1}$

$\dfrac{{W}_{2}}{50} + 1 = \dfrac{{W}_{2}}{50 \times 0.75}$

${W}_{2} = 150g$