Mole ConceptHard
Question
The density of gold is 19.7 g/cm3. The radius of gold atom is [Au = 197, NA = 6 × 1023, (10π)1/3 = 3.15]
Options
A.1.587 × 10–8 m
B.1.587 × 10–9 m
C.1.587 × 10–10 m
D.1.587 × 10–12 m
Solution
$\frac{4}{3}\pi r^{3} = \frac{197}{6 \times 10^{23} \times 19.7}\text{c}\text{m}^{3} \Rightarrow r = 1.587 \times 10^{- 8}\text{cm}$
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