Question

The mass of $3.01\times {10}^{23}$ molecules of a gas iss $32.0g$. What volume does $8.00$ g of the gas occupy at STP?

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

Answer 1

The correct option is A $2.80L$

Mass of $3.01$ $\times$ ${10}^{23}$ molecules is $32$ grams.

Mass of $1$ molecule is $\frac{32}{3.01}$ $\times$ $\frac{1}{{10}^{23}}$ grams.

Thus, mass of $6.023$ $\times$ ${10}^{23}$ molecules is $\frac{32}{3.01}$ $\times$ $\frac{1}{{10}^{23}}$ $\times$ $6.023$ $\times$ ${10}^{23}$ grams, which is almost equal to $64$ grams.

Thus, 64 grams is the molecular weight of the gas.

One mole of any gas occupy $22.4$ $L$ at STP.

64 grams occupy $22.4$ $L$ at STP.

1 gram occupy $\frac{22.4}{64}$ $L$ at STP.

Thus, 8 grams will occupy $\frac{22.4}{64}$ $\times$ $8$ $L$ at STP, which is equal to 2.80 L.

Thus, option A is the correct answer.