Molar mass of octane = 114 g/mol.
From the lowering of vapour pressure, we have,
PΔP=M2W2+M1W1M2W2
Where W2andM2 are mass and molar mass of solute
and W1andM1 are mass and molar mass of octane.
10075=50g/molW2+114g/mol114g50g/molW2
0.75=50W2+150W2
50W2+1=50×0.75W2
W2=150g