Question

1. a sample contains (1.20 \times 10^{24}) atoms of magnesium. what is the mass of this sample?

○ 60.0 g

○ 12.0 g

○ 24.3 g

○ 48.6 g

clear all

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 Mg is \( 24.3\space g/mol\). We need to find the moles of Mg first from the number of atoms, then find the mass.

Step2: Calculate moles of Mg

Moles (\(n\)) = \(\frac{\text{Number of atoms}}{\text{Avogadro's number}}\)
Given number of atoms = \( 1.20\times 10^{24}\) atoms
\(n=\frac{1.20\times 10^{24}\space \text{atoms}}{6.022\times 10^{23}\space \text{atoms/mol}}\approx 1.993\space mol\) (approx 2 mol for simplicity, but let's do precise)

Step3: Calculate mass of Mg

Mass (\(m\)) = \(n\times M\), where \(M = 24.3\space g/mol\)
\(m = 1.993\space mol\times24.3\space g/mol\approx 48.4\space g\) (close to 48.6 g, considering more precise Avogadro's number or rounding) Wait, let's recalculate:

Wait, \(1.20\times 10^{24}\div6.022\times 10^{23}= \frac{1.20}{6.022}\times 10^{24 - 23}=\frac{1.20}{6.022}\times 10\approx 1.993\) mol (approx 2 mol? Wait no: 6.022e23 is 1 mol, so 1.20e24 is (1.20e24)/(6.022e23)= 1.2010/6.022 ≈ 1.993 mol. Then mass = 1.993 mol 24.3 g/mol ≈ 1.99324.3. Let's compute 224.3=48.6, and 1.993 is almost 2, so 48.6 g.

Wait, maybe using Avogadro's number as 6.02×10²³ for simplicity:

\(n=\frac{1.20\times 10^{24}}{6.02\times 10^{23}}=\frac{1.20}{6.02}\times 10=\frac{12}{6.02}\approx 1.993\space mol\) (≈2 mol)

Then mass = 2 mol *24.3 g/mol = 48.6 g. So the answer is 48.6 g.

Answer:

48.6 g (the option with 48.6 g)