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

Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).
(Rev. 27-Mar-2023)

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Question 1.29 NCERT Class XI Chemistry

Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Video Explanation(detailed solution given after this video)

Please watch this youtube video for a quick explanation of the solution:

Solution in Detail

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

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