Question

The mass of $6.02\times {10}^{23}$ molecules of a gas is 64.0 grams. What volume does 8.00 grams of the gas occupy at standard temperature and pressure?

• A. $2.80$ $L$
• B. $8.00$ $L$
• C. $11.2$ $L$
• D. $22.4$ $L$
• E. $64.0$ $L$

Answer 1

Answer: (A)

$2.80$ $L$

$6.02\times {10}^{23}$ is Avogadro's number of molecules. It corresponds to 1 mole of gas.
The mass of 1 mole of gas is 64.0 grams.
8.00 grams of gas corresponds to $\frac{8.00g}{64.0g/mol}=\frac{1}{8.0}moles$
At standard temperature and pressure, 1 mole of a gas occupies a volume of 22.4 $L$.
So $\frac{1}{8.0}moles$ of a gas will occupy a volume of $22.4L/mol\times \frac{1}{8.0}moles=2.80L$