What is the molar mass of a gas if 8.482 grams of the gas will occupy 2.500 liters of space at STP?

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What is the molar mass of a gas if 8.482 grams of the gas will occupy 2.500 liters of space at STP?

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Answer 1 · 2016-02-10T19:56:24

Approx $70 - 80$ $70-80$ $g \cdot m o l^{- 1}$ $g*mol^-1$ .

Explanation:

At STP $1$ $1$ mole of ideal gas occupies a volume of $22.41$ $22.41$ $L$ $L$ .

We assume that the unknown gas behaves ideally, so that the molar quantity is $\frac{2.500 \cdot L}{22.41 \cdot L \cdot m o l^{- 1}}$ $(2.500*L)/(22.41*L*mol^-1)$ $=$ $=$ $0.1116$ $0.1116$ $m o l$ $mol$ .

But this molar quantity has a mass of $8.482$ $8.482$ $g$ $g$ .

So $Molar mass$ $"Molar mass"$ $=$ $=$ $\frac{8.482 \cdot g}{0.1116 \cdot m o l}$ $(8.482*g)/(0.1116*mol)$ $=$ $=$ $76.00$ $76.00$ $g \cdot m o l^{- 1}$ $g*mol^-1$ .

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What is the molar mass of a gas if 8.482 grams of the gas will occupy 2.500 liters of space at STP?

1 Answer

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