Discussion
How many Nickel atoms are present in 25 g Ni(MW of Nickel is 58.71 g) ?
Answer: C
25 g of nickel contains about 2.57 × 1023 atoms.
Divide the mass by the molar mass to get moles, then multiply by Avogadro's number:
\[ \begin{aligned} n &= \frac{25}{58.71} = 0.426\ \text{mol} \\ N &= 0.426 \times 6.022 \times 10^{23} \\ &\approx 2.56 \times 10^{23} \end{aligned} \]
This rounds to the 2.57 × 1023 of the key. The other options are slips of the decimal point or digits: 25.7 and 52.7 × 1023 are ten and twenty times too many atoms, which would need about 250 g or 500 g of nickel, and 5.27 × 1023 would need about 51 g.
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