Question

The mass of $112c{m}^3$ of ${CH}_4$ gas at STP is :

• A. 0.08g
• B. 1.6g
• C. 0.16g
• D. 0.8g

Answer 1

The correct option is A 0.08g

$1L=1000c{m}^3$

so, volume = $\frac{112}{1000}=0.112L$

Molar mass of $C{H}_4$ - $12+4=16$

At STP, 1 mole = $22.4L$

It means $22.4L$ contains $16g$ methane.

$0.112L$ contains = $\frac{16}{22.4}\times 0.112=0.08g$

Hence, option A is the correct answer.