Question

how many molecules of o₂ are present in a 3.90 l flask at a temperature of 273 k and a pressure of 1.00 atm?

• use 6.022 × 10²³ mol⁻¹ for avogadros number.

select the correct answer below:
○ 1.05 × 10²³ molecules
○ 1.74 × 10²³ molecules
○ 3.74 × 10²⁴ molecules
○ 9.60 × 10²² molecules

Explanation:

Step1: Use the ideal gas law \(PV = nRT\)

We know \(P = 1.00\text{ atm}\), \(V=3.90\text{ L}\), \(T = 273\text{ K}\), and \(R=0.0821\text{ L}\cdot\text{atm/mol}\cdot\text{K}\).
Rearrange for \(n\): \(n=\frac{PV}{RT}\)
Substitute the values: \(n=\frac{1.00\times3.90}{0.0821\times273}\)
Calculate \(n=\frac{3.90}{22.4133}\approx0.174\text{ mol}\)

Step2: Use Avogadro's number \(N = n\times N_A\)

Given \(N_A = 6.022\times 10^{23}\text{ mol}^{-1}\)
\(N=0.174\times6.022\times 10^{23}\)
\(N=(0.174\times6.022)\times 10^{23}\)
\(0.174\times6.022 = 1.047828\approx1.05\)
So \(N = 1.05\times 10^{23}\text{ molecules}\)

Answer:

\(1.05\times 10^{23}\text{ molecules}\)