The mass of non-volatile, non-electrolyte solute `("molar mass" = 50 g "mol"^(-1))` needed to be dissolved in `114 g` octane to reduce its vapour pressure to 75%, is :

Options

A. 37.5g
B. 75g
C. 150 g
D. 50 g

Answer

C

Explanation

Molar mass of octane = 144 g/mol
From the lowering vapour pressure we have,
`(DeltaP)/P =((W_2)/(M_2))/(W_2/M_2 + W_1/M_1) `
`75/100 = (W_2/(50g//mol))/(w_2/(50g//mol)+(114g)/(114 g//mol))`
`0.75 = (W_2/50)/(W_2/50 + 1)`
`W_2/50 +1 = W_2/(50xx0.75)`
`W_2 = 150g`
Note : `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.