Question

The volume occupied by $8.8gC{O}_2$ at $31{}^{o}C$ and 1 atm pressure is

• A. 5 ml
• B. 0.2 L
• C. 5 L
• D. 2 L

Answer 1

Answer: (C)

Here: Given mass of $C{O}_2=8.8\text{gram}$

Molar mass of $C{O}_2=44\text{gram}$

Now, number of moles of $C{O}_2=\frac{8.8}{44}=0.2\text{moles}$

$n=2$

Using $PV=nRT$

$V=\frac{nRT}{P}=$ volume occupied.

$n=0.2,R=\frac{0.0821L-atm}{Kmol}$

$T=31{}^{o}C=31+273=304K$

$P=1\text{atm}$

$\text{Volume}=\frac{0.2mol\times 0.0821\times L-atm\times 304K}{1atm\times Kmol}$

$=0.2\times 0.0821\times 304L$

$=4.99$

Approx. $5$ Liters