Question

what volume will 3.01 x 10^{23} molecules of gas at stp occupy? 44.8 l 22.4 l 11.2 l 18.8 l

Explanation:

Step1: Calculate the number of moles

We know that \(1\) mole of any substance contains \(N_{A}=6.02\times 10^{23}\) molecules.
Let \(n\) be the number of moles of the gas. Given \(N = 3.01\times 10^{23}\) molecules.
Using the formula \(n=\frac{N}{N_{A}}\), we substitute \(N = 3.01\times 10^{23}\) and \(N_{A}=6.02\times 10^{23}\)
\(n=\frac{3.01\times 10^{23}}{6.02\times 10^{23}}=\frac{1}{2}= 0.5\) mol.

Step2: Use the molar volume at STP

At STP (Standard Temperature and Pressure: \(T = 273\) K and \(P=1\) atm), the molar volume of a gas \(V_{m}=22.4\) L/mol.
The volume \(V\) of a gas is given by the formula \(V=n\times V_{m}\).
Substituting \(n = 0.5\) mol and \(V_{m}=22.4\) L/mol.
\(V=0.5\times22.4 = 11.2\) L.

Answer:

C. 11.2 L