A $25.0 \mathrm{~mm} \times 40.0 \mathrm{~mm}$ piece of gold foil is $0.25 \mathrm{~mm}$ thick. The density…
- $7.7 \times 10^{23}$
- $1.5 \times 10^{23}$
- $4.3 \times 10^{21}$
- $1.47 \times 10^{22}$
Solution
$=250 \times 10^{-3} \mathrm{~cm}^{3}$
Mass of gold foil $=19.32 \times 250 \times 10^{-3} \mathrm{~g}$
$=4.83 \mathrm{~g}$
No. of gold atoms $=\frac{4.83}{197} \times \mathrm{N}_{\mathrm{A}}$
$=1.47 \times 10^{22}$ /
Asked in: JEE-TOPICTESTS-CHEMISTRY
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