Use the data given in the following table to calculate the molar mass of naturally occurring argon having 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 %

Question

Use the data given in the following table to calculate the molar mass of naturally occurring argon having 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 % (A) 89.94 g mol-1 (B) 39.95 g mol-1 (C) 94.39 g mol-1 (D) 49.39 g mol-1

Tardigrade Admin · Accepted Answer

Molar mass of naturally occurring argon = (35.96755 × 0.337 + 37.96272 × 0.063 + 39.9624 × 99.600/0.337 + 0.063 + 99.600) = (12.12 + 2.39 + 3980.255/100) = (3994.765/100) = 39.95 g mol-1

Isotope	Isotopic molar mass	Abundance
$^{36} A r$	$35.96755 g m o l^{- 1}$	$0.337%$
$^{38} A r$	$37.96272 g m o l^{- 1}$	$0.063%$
$^{40} A r$	$39.9624 g m o l^{- 1}$	$99.600%$

Q. Use the data given in the following table to calculate the molar mass of naturally occurring argon having isotopes:

Isotope Isotopic molar mass Abundance

$^{36} A r$ $35.96755 g m o l^{- 1}$ $0.337%$

$^{38} A r$ $37.96272 g m o l^{- 1}$ $0.063%$

$^{40} A r$ $39.9624 g m o l^{- 1}$ $99.600%$

$89.94 g m o l^{- 1}$

$39.95 g m o l^{- 1}$

$94.39 g m o l^{- 1}$

$49.39 g m o l^{- 1}$

Molar mass of naturally occurring argon
$= \frac{35.96755 \times 0.337 + 37.96272 \times 0.063 + 39.9624 \times 99.600}{0.337 + 0.063 + 99.600}$
$= \frac{12.12 + 2.39 + 3980.255}{100}$
$= \frac{3994.765}{100}$
$= 39.95 g m o l^{- 1}$

Q. Use the data given in the following table to calculate the molar mass of naturally occurring argon having isotopes: Isotope Isotopic molar mass Abundance 36Ar 35.96755gmol−1 0.337% 38Ar 37.96272gmol−1 0.063% 40Ar 39.9624gmol−1 99.600%

89.94gmol−1

39.95gmol−1

94.39gmol−1

49.39gmol−1