Question

16.0 g of NaOH is present in 100 ml of an aqueous solution. Its density is 1.06 g/ml. Mole fraction of the solute is approximately:

• A. $25/27$
• B. $2/27$
• C. $1/27$
• D. $26/27$

Answer 1

The correct option is B $2/27$

Moles of NaOH $=\frac{16}{40}$ $=0.4$

Density = 1.06 g/ml

1ml weighs 1.06 gms

100 ml solution weighs $\frac{1.06}{1}\times 100=106gms$

Therefore, amount of solvent $=$ 106 $-$ 16 $=$ 90 gms

Moles of solvent $=\frac{90}{18}=5$

Now,

$X_{sol}=$ Mole fraction of solute

$=\frac{Molesofsolute}{Molesofsolute+molesofsolvent}$

$X_{sol}=\frac{0.4}{0.4+5}=\frac{2}{27}$