(solved)Question 1.29 of NCERT Class XI Chemistry Chapter 1

Let mols of ethanol = $\displaystyle n_E$

Let mols of water = $\displaystyle n_w$

by def. mol fraction of ethanol $\displaystyle \frac{n_E}{n_E + n_w}$

$\displaystyle \underline{\underline{\text{Step 1: Calculate Mols of Water}}}$

Given $\displaystyle \frac{n_E}{n_E + n_w}$ = 0.040

$\displaystyle \therefore \frac{n_E/n_w}{n_E/n_w + 1}$ = 0.040

$\displaystyle \implies \frac{n_E}{n_w} = \frac{1}{24} \text{. . . (1)}$

Hence, solution is mainly of water

Let volume of solution = 1 liters

given, density of water = 1 kg/liter

we know mass = vol x density

$\displaystyle \implies$mass of water = 1 x 1 (kg/liter) = 1000 grams

$\displaystyle \implies n_w $= mass/Molar Mass = 1000/18 = (500/9)

$\displaystyle \underline{\underline{\text{Step 2: Calculate Mols of Ethanol}}}$

By (1): $\displaystyle \: \frac{n_E}{n_w} = 1/24$

$\displaystyle \implies n_E = \frac{1}{24} \times \frac{500}{9}= 2.31 \text{ mol}$

$\displaystyle \underline{\underline{\text{Lastly: Calculate Molarity of Ethanol}}}$

Molarity $\displaystyle = \frac{\text{no. of moles of solute}}{\text{vol. of soln. in liters}}$

$\displaystyle = \bigg(\frac{2.31\text{ mols}}{1 \text{ liter}}\bigg) = 2.31 M \: \underline{ Ans}$