Question

how many molecules of nitrogen gas can be present in a 2.5 l flask at 323 k and 650 mm hg? 3.1 x 10^{23} molecules 3.6 x 10^{25} molecules 0.081 molecules 4.9 x 10^{22} molecules 2.1 x 10^{-23} 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 \(PV = nRT\)

Where \(V = 2.5\ L\), \(R=0.0821\ L\cdot atm/(mol\cdot K)\), \(T = 323\ K\). Rearranging for \(n\) (moles), we get \(n=\frac{PV}{RT}\)
Substitute 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 \(N = n\times N_A\) (\(N_A=6.022\times 10^{23}\ mol^{-1}\))
\(N=0.0806\times6.022\times 10^{23}\)
\(N\approx4.85\times 10^{22}\approx4.9\times 10^{22}\)

Answer:

\(4.9\times 10^{22}\ molecules\)