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

asked Apr 21, 2023 71.1k views

2 votes

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6 votes

We know,

Molar mass of gold is 196.966 g

Thus,

Required mass is:

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

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