Question

Which of the following statements are correct?

(i) Mole fraction of a solute + mole fraction of solvent $=$ $1$

(ii) If equal weights of helium and methane are present in a gaseous mixture, then the mole fraction of He is $\frac{4}{5}$

(iii) The mole fraction of water in the aqueous solution of NaOH is $0.8$. The molality of the solution is nearly 14 moles kg${}^{-1}$

• A. (i) and (ii)
• B. (ii) and (iii)
• C. (i) and (iii)
• D. All are correct

Answer 1

The correct option is D All are correct

(i) $X_A=\frac{n_A}{n_A+n_B},X_B=\frac{n_B}{n_A+n_B}$
Hence, $X_A+X_B=1$

(ii) $X_{He}=\frac{\frac{X}{4}}{\frac{X}{4}+\frac{X}{16}}=0.80=\frac{4}{5}$.
(iii) Mole fraction of solvent = $0.8$

We know that

$Molality=\frac{\text{No. of moles of solute}}{\text{Mass of solvent in Kg}}$

Let the mass of solvent = $\text{1000 grams}$

$\text{No. of moles of solvent}=\frac{\text{1000}}{\text{18}}$ = $55.55$

$x_{solute}=1-x_{solvent}$ = $1-0.8=0.2$

$Molality=\frac{x_{solute}}{x_{solvent}}X\text{N}o.of{H}_{2}O$

= $\frac{0.2}{0.8}X55.55=13.88mol/kg=14mol/kg$