Question

a balloon contains ( 4.50 \times 10^{22} ) atoms of helium (he) gas. calculate the mass of helium in grams.
( \bigcirc ) 0.0185 g
( \bigcirc ) 0.299 g
( \bigcirc ) 0.185 g
( \bigcirc ) 0.0747 g

Explanation:

Step1: Recall Avogadro's number and molar mass

Avogadro's number is \( 6.022\times 10^{23}\) atoms/mol, and the molar mass of He is \( 4.00\) g/mol (since atomic mass of He is approximately 4.00 amu). We need to find the moles of He first using the number of atoms and then convert moles to mass.

Step2: Calculate moles of He

Moles (\(n\)) = \(\frac{\text{Number of atoms}}{\text{Avogadro's number}}\)
Given number of atoms = \( 4.50\times 10^{22}\)
So, \(n=\frac{4.50\times 10^{22}}{6.022\times 10^{23}\text{ atoms/mol}}\)
\(n=\frac{4.50}{60.22}\text{ mol}\) (since \(10^{22}/10^{23}=10^{-1} = 0.1\), so \(4.50\times 10^{22}/6.022\times 10^{23}=4.50/(6.022\times 10)=4.50/60.22\approx0.0747\text{ mol}\) Wait, no, wait: \(6.022\times 10^{23}\) is Avogadro's number. So \(4.50\times 10^{22}\) atoms \(\times\frac{1\text{ mol}}{6.022\times 10^{23}\text{ atoms}}\)
\(=\frac{4.50\times 10^{22}}{6.022\times 10^{23}}\text{ mol}=\frac{4.50}{6.022\times 10}\text{ mol}=\frac{4.50}{60.22}\text{ mol}\approx0.0747\text{ mol}\)? Wait, no, 4.50e22 divided by 6.022e23: 4.50/60.22 ≈ 0.0747? Wait, 6.022e23 is 60.22e22, so 4.50e22 /60.22e22 = 4.50/60.22 ≈ 0.0747 mol. Then mass = moles × molar mass. Molar mass of He is 4.00 g/mol. So mass = 0.0747 mol × 4.00 g/mol ≈ 0.299 g? Wait, no, wait, let's recalculate:

Wait, \(4.50\times 10^{22}\) atoms. Avogadro's number is \(6.022\times 10^{23}\) atoms/mol. So moles \(n=\frac{4.50\times 10^{22}}{6.022\times 10^{23}}=\frac{4.50}{6.022\times 10}=\frac{4.50}{60.22}\approx0.0747\) mol? Wait, no, 4.50e22 /6.022e23 = (4.50/6.022) 10^(22-23) = 0.747 10^(-1) = 0.0747 mol? Wait, no, 4.50 divided by 6.022 is approximately 0.747, then times 10^-1 is 0.0747? Wait, no, 4.50e22 is 0.450e23, so 0.450e23 /6.022e23 = 0.450/6.022 ≈ 0.0747 mol. Then mass = 0.0747 mol * 4.00 g/mol = 0.2988 g ≈ 0.299 g.

Wait, let's do it step by step:

1. Number of atoms = \(4.50\times 10^{22}\)
2. Avogadro's number (\(N_A\)) = \(6.022\times 10^{23}\) atoms/mol
3. Molar mass of He (\(M\)) = 4.00 g/mol

First, calculate moles (\(n\)):

\(n=\frac{\text{Number of atoms}}{N_A}=\frac{4.50\times 10^{22}\text{ atoms}}{6.022\times 10^{23}\text{ atoms/mol}}\)

Simplify the exponents: \(10^{22}/10^{23}=10^{-1}\), so:

\(n=\frac{4.50}{6.022}\times 10^{-1}\text{ mol}\)

Calculate \(4.50/6.022\approx0.747\), so:

\(n\approx0.747\times 10^{-1}\text{ mol}=0.0747\text{ mol}\)

Then, calculate mass (\(m\)) using \(m = n\times M\):

\(m = 0.0747\text{ mol}\times 4.00\text{ g/mol}\)

\(m\approx0.2988\text{ g}\approx0.299\text{ g}\)

Answer:

0.299 g (corresponding to the option "0.299 g")