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What is the mass, in grams, of 1.34 × 10^18 atoms of gold present in one nanoparticle?
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15 votes
15 votes
What is the mass, in grams, of 1.34 × 10^18 atoms of gold present in one nanoparticle?

User Sachinrahulsourav
by
2.3k points

1 Answer

15 votes
15 votes

We know,

Molar mass of gold is 196.966 g

Thus,

Required mass is:


\begin{gathered} m=1.34*(10^(18))/(6.023*10^(23))*196.966g \\ \Rightarrow m=0.00044g \end{gathered}

User Rotsen
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3.3k points
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