Question

$\text{The volume of}2.8\text{g}\text{of carbon monoxide at}{27}^{\circ }\text{C}\text{and}0.821\text{atm is}$

• A. $30\text{L}$
• B. $3\text{L}$
• C. $0.3\text{L}$
• D. $1.5\text{L}$

Answer 1

The correct option is B $3\text{L}$

According to the ideal gas equation, we have
$PV=nRT$
$\Rightarrow PV=\left(\frac{w}{M}\right)RT$
$\Rightarrow V=\left(\frac{w}{M}\right)\left(\frac{RT}{P}\right)$
Given values are:
$w$ = $2.8\text{g}$
$M$ = Molar mass of $CO$ = $28{\text{g mol}}^{{}^{-1}}$
$T$ = ${27}^{\circ }\text{C}$ = $(273+27)\text{K}$ = $300\text{K}$
$P$ = $0.821\text{atm}$
$R=0.0821{\text{L atm mol}}^{-1}{\text{K}}^{-1}$
Putting the values in the formula we get :
$V=\left(\frac{2.8\text{g}}{28\text{g}}\right){\text{mol}}^{-1}\times \frac{\left(0.0821{\text{L atm mol}}^{-1}{\text{K}}^{-1}\right)\times \left(300\text{K}\right)}{(0.821\text{atm)}}$
$=3\text{L}$