Question

The volume occupied by $88g$ of $C{O}_2$ at ${30}^{\circ }C$ and $1bar$ pressure will be:

• A. $5.05L$
• B. $49.8L$
• C. $2L$
• D. $55L$

Answer 1

Answer: (B)

$49.8L$

$w=88g,M=44gmo{l}^{-1},T=273+30=303K$
$P=1bar,V=?$
$PV=nRT$ or $V=\frac{nRT}{P}=\frac{w}{m}\frac{RT}{P}$
$V=\frac{88\times 0.0821\times 303}{44\times 1}=49.8L$.