4.0 g of NaOH are contained in one decilitre of solution. Calculate t…

The correct answer is Option (1) → $0.018$ ##

$\text{Density of solution} = 1.038\text{ g/cm}^3$

$\text{Mass of } 100\text{ cm}^3\text{ of solution} = 1.038 \times 100$

$= 1038\text{ g}$

$\text{Mass of NaOH in } 100\text{ cm}^3\text{ solution} = 4\text{ g}$

$\text{Mass of water} = 103.8 - 4.0 = 99.8\text{ g} $

$\text{Number of moles of NaOH } (n_{\text{A}}) = \frac{4.0}{40}$

$= 0.1\text{ mol}$

$\text{Number of moles of water } (n_{\text{B}}) = \frac{99.8}{18}$

$= 5.54\text{ mol}$

$∴\text{Mole fraction of NaOH } (\chi_{\text{A}}) = \frac{n_{\text{A}}}{n_{\text{A}} + n_{\text{B}}}$

$= \frac{0.1}{0.1 + 5.54}$

$= 0.018 $