A $25.0 \mathrm{~mm} \times 40.0 \mathrm{~mm}$ piece of gold foil is $0.25 \mathrm{~mm}$ thick. The density…

A $25.0 \mathrm{~mm} \times 40.0 \mathrm{~mm}$ piece of gold foil is $0.25 \mathrm{~mm}$ thick. The density of gold is $19.32 \mathrm{~g} / \mathrm{cm}^{3}$. How many gold atoms are in the sheet? (Atomic weight $: \mathrm{Au}=197.0$ )
  1. $7.7 \times 10^{23}$
  2. $1.5 \times 10^{23}$
  3. $4.3 \times 10^{21}$
  4. $1.47 \times 10^{22}$

Solution

Volume of gold foil $=25 \times 40 \times 0.25 \mathrm{~mm}^{3}$
$=250 \times 10^{-3} \mathrm{~cm}^{3}$
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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