Question

part a
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 approximately \( 238.03 \, \text{g/mol} \), and Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \).

Step2: Use formula \( m=\frac{N\times M}{N_A} \)

Given \( N = 1.27 \times 10^{20} \) atoms, \( M = 238.03 \, \text{g/mol} \), \( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \).
Substitute into the formula:
\( m=\frac{1.27\times 10^{20} \times 238.03}{6.022\times 10^{23}} \)

Step3: Calculate the value

First, multiply the numerator: \( 1.27\times 10^{20} \times 238.03 \approx 1.27\times238.03\times 10^{20} \approx 302.2981\times 10^{20} = 3.022981\times 10^{22} \)
Then divide by \( 6.022\times 10^{23} \):
\( m=\frac{3.022981\times 10^{22}}{6.022\times 10^{23}} \approx \frac{3.022981}{60.22} \approx 0.0502 \, \text{g} \)

Answer:

\( 0.0502 \) (or more precise calculation: \( \frac{1.27\times10^{20}\times238.03}{6.022\times10^{23}} \approx 0.0502 \, \text{g} \))