### Question 1.32

NCERT Class XI Chemistry

Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes: Isotope Isotopic molar mass Abundance 36Ar 35.96755 g mol-1 0.337% 38Ar 37.96272 g mol-1 0.063% 40Ar 39.9624 g mol-1 99.600%

### Video Explanation

(detailed solution given after this video)

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

### Solution in Detail

Take 100 grams of natural argon

by abundance, 36Ar = 0.337% = 0.337 g

Similarly, 38Ar = 0.063 g, 40Ar = 99.600 g

no. of moles formula $\displaystyle n = \frac{\text{mass}}{\text{Molar Mass}}$

$\displaystyle \underline{\underline{\text{Step 1: moles of }\:^{36}\text{Ar}}}$

$\displaystyle n = \frac{0.337 \text{ g}}{35.96755 \text{ g/mol}} = 0.00937 \text{ mol}$

$\displaystyle \underline{\underline{\text{Step 2: moles of }\:^{38}\text{Ar}}}$

$\displaystyle n = \frac{0.063 \text{ g}}{37.96272 \text{ g/mol}} = 0.0016 \text{ mol}$

$\displaystyle \underline{\underline{\text{Step 3: moles of }\:^{40}\text{Ar}}}$

$\displaystyle n = \frac{99.600 \text{ g}}{39.9624 \text{ g/mol}} = 2.4923 \text{ mol}$

$\displaystyle \underline{\underline{\text{Finally: Combined Molar Mass}}}$

Combined molar mass = $\displaystyle \frac{\text{combined mass}}{\text{combined moles}}$

$\displaystyle =\frac{100 \text{ g}}{0.00937 + 0.0016+2.4923 }$

$\displaystyle = 39.948 \text{ g/mol}\:\underline{ Ans}$

### Similar Posts

This Blog Post/Article "(solved)Question 1.32 of NCERT Class XI Chemistry Chapter 1" by Parveen is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.