# Question #e96d5

Sep 7, 2017

$0.932 \times {10}^{- 5}$

#### Explanation:

It's so easy to understand.

The only thing we have to know is that 1 mol of any gas has a volume of 24.04 L at NTP (Normal Temperature and Pressure)

Here the volume is given as 2.24 mL.

1 L = 1000 mL

Hence $\text{1 mL" = 10^(-3) " L}$

$\text{2.24 mL" = 2.24 xx 10^(-3) " L}$

${\text{24.04 L of CO"_2 = "1 mol of CO}}_{2}$ (at NTP only)

$2.24 \times {10}^{- 3} {\text{ L CO"_2 xx ("1 mol CO"_2)/("24.04 L CO"_2) = 9.32 xx 10^(-5) " mol CO}}_{2}$

Therefore 2.24 mL ${\text{CO}}_{2}$ at NTP contains $0.932 \times {10}^{- 5} {\text{ mol CO}}_{2}$.