Question

1.27×10²⁰ uranium atoms
m =
g
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Explanation:

Step1: Recall molar mass of U

Uranium (U) has a molar mass of \( 238.03 \, \text{g/mol} \) (from periodic table). Avogadro's number is \( 6.022 \times 10^{23} \, \text{atoms/mol} \).

Step2: Use formula \( m = \frac{\text{atoms} \times \text{molar mass}}{\text{Avogadro's number}} \)

Substitute values: \( \text{atoms} = 1.27 \times 10^{20} \), \( \text{molar mass} = 238.03 \, \text{g/mol} \), \( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \).
\( m = \frac{1.27 \times 10^{20} \times 238.03}{6.022 \times 10^{23}} \)
Calculate numerator: \( 1.27 \times 10^{20} \times 238.03 \approx 1.27 \times 238.03 \times 10^{20} \approx 302.3 \times 10^{20} = 3.023 \times 10^{22} \)
Divide by denominator: \( \frac{3.023 \times 10^{22}}{6.022 \times 10^{23}} = \frac{3.023}{60.22} \approx 0.0502 \, \text{g} \)

Answer:

\( 0.0502 \) (or more precise: \( \approx 0.0501 \, \text{g} \) if calculated with more precision)