Question

how many molecules of nitrogen gas can be present in a 2.5 l flask at 323 k and 650 mm hg?
2.1 x 10^23 molecules
4.9 x 10^22 molecules
3.6 x 10^25 molecules
3.1 x 10^23 molecules
0.081 molecules

Explanation:

Step1: Convert pressure to atm

We know that \(1\ atm = 760\ mmHg\). So, \(P=\frac{650}{760}\ atm\approx0.855\ atm\)

Step2: Use ideal gas law \(PV = nRT\)

Given \(V = 2.5\ L\), \(T=323\ K\), \(R = 0.0821\ L\cdot atm/(mol\cdot K)\). Rearranging for \(n\) (number of moles), we get \(n=\frac{PV}{RT}\)
Substitute the values: \(n=\frac{0.855\times2.5}{0.0821\times323}\)

$$n=\frac{2.1375}{26.5183}\approx0.0806\ mol$$

Step3: Calculate number of molecules

Using \(N = n\times N_A\) (\(N_A=6.022\times 10^{23}\ molecules/mol\))
\(N=0.0806\times6.022\times 10^{23}\)

$$N=(0.0806\times6.022)\times10^{23}\approx 4.85\times 10^{22}\approx4.9\times 10^{22}\ molecules$$

Answer:

\(4.9\times 10^{22}\ molecules\) (the second option)