Question

Calculate the weight of a non-volatile solute $(mol./wt.40)$ which should be dissolved in $114g$ octane to reduce its vapour pressure to $80$%.

Answer 1

Weight of non volatile solute m

mol $wt.M=40$

octane $(C_8{H}_{18}=8\times 12+18=114g)$

No. of mole = $\frac{114}{114}=1$ mole

Number of mole of octane = 1 mole

Number of mole of solute = $\frac{m}{40}$

mole fraction = $\frac{1}{1+\frac{m}{40}}=\frac{40}{40+m}$

$P_A^o\frac{40}{40+m}=\frac{P_A^o\times 80}{100}$

${50}^o=40+m$

$m+10g$