Answer to Question #113434 in Physical Chemistry for Charan

Question

Active mass of 5.6L of N2 at STP

Answer 1

According to the definition, the active mass of nitrogen is:

$c = \frac{n}{V}$ ,

where $n$ is the number of the moles of nitrogen and $V$ is the volume of the system. It is the same as molar concentration. Let's find the number of the moles of nitrogen in 5.6 L at STP (pressure $p$ =10⁵ Pa, temperature $T$= 273.15 K) . Assuming that nitrogen has the behavior of the ideal gas:

$pV = nRT$

$n = \frac{pV}{RT}$ ,

where $R$ is the ideal gas constant, equal to 8.314·10³ L Pa / (K mol).

$n = \frac{10^5·5.6}{8.314 ·10^3·273.15} = 0.247$ mol

Finally, the active mass of nitrogen is:

$c = \frac{0.247}{5.6} = 0.044$ mol/L

Answer: Active mass of 5.6L of N2 at STP is 0.044 mol/L