### Question 1.9

NCERT Class XI Chemistry

Calculate the atomic mass (average) of chlorine using the following data: % natural abundance of 35Cl is 75.77% and molar mass is 34.9689 g/mol. The % natural abundance of 37Cl is 24.23% and its molar mass is 36.9659 g/mol.

### Solution in Brief

Molar mass will be the weighted average. Following is the detailed explanation:

Take 100 mols of the chlorine mixture. This sample will have 75.77 mols of 35Cl and their mass would be (75.77 x 34.9689 grams). Similarly, the mass of 37Cl will be (24.23 x 36.9659 grams). The combined total mass of 100 mols is (75.77 x 34.9689) + (24.23 x 36.9659) = 3545 grams, correct to 4 significant figures. Hence, the mass of 1 mol will be 3545/100 = 35.45 g, which is the molar mass 35.45 g/mol answer!

### Video Explanation

(detailed solution given after this video)

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

### Solution in Detail

$ \begin{aligned} &\text{take } 100 \text{ mol of chlorine}\\\\ &\text{moles of } ^{35}\text{Cl } = 75.77 \text{ mol}\\\\ &\text{mass of } ^{35}\text{Cl } = 75.77 \times 34.9689\text{ g . . .(1)}\\\\ &\text{moles of } ^{37}\text{Cl } = 24.23 \text{ mol}\\\\ &\text{mass of } ^{37}\text{Cl } = 24.23 \times 36.9659\text{ g . . .(2)}\\\\ &\therefore \text{mass of 100 moles} = \text{(1) + (2)}\\\\ &\therefore \text{mass of 1 mole} = \frac{\text{(1) + (2)}}{100}\\\\ &= \frac{(75.77 \times 34.9689) + (24.23 \times 36.9659) }{100}\\\\ &= \frac{ 3545 \text{ (4 SF)} }{100} =35.45 \text{ g/mol}\:\underline{Ans} \end{aligned} $

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