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Calculate the number of atoms in $39.4 \mathrm{~g}$ gold. Molar mass of gold is $197 \mathrm{~g} \mathrm{~mole}^{-1}$.
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Verified Answer
As given that, molar mass of gold $=197 \mathrm{~g} / \mathrm{mol}$
Molar Mass i.e. mass of Avogadro's number of atoms $=6.023 \times 10^{23}$ atoms
So, $197 \mathrm{gm}$ gold contains $=6.023 \times 10^{23}$ atoms
Number of atoms $1 \mathrm{gm}$ gold contain
$=\frac{6.023 \times 10^{23}}{197}$ atoms
Number of gold atoms $39.4 \mathrm{gm}$ gold contains
$$
\begin{aligned}
&=\frac{6.023 \times 10^{23} \times 39.4}{197} \\
&=\frac{12.046 \times 10^{23}}{10}=1.2046 \times 10^{23} \text { atoms }
\end{aligned}
$$
Number of gold atoms.
Molar Mass i.e. mass of Avogadro's number of atoms $=6.023 \times 10^{23}$ atoms
So, $197 \mathrm{gm}$ gold contains $=6.023 \times 10^{23}$ atoms
Number of atoms $1 \mathrm{gm}$ gold contain
$=\frac{6.023 \times 10^{23}}{197}$ atoms
Number of gold atoms $39.4 \mathrm{gm}$ gold contains
$$
\begin{aligned}
&=\frac{6.023 \times 10^{23} \times 39.4}{197} \\
&=\frac{12.046 \times 10^{23}}{10}=1.2046 \times 10^{23} \text { atoms }
\end{aligned}
$$
Number of gold atoms.
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