Question

The density of gold is $19 \,g/cm^3$. If $1.9 \times 10^{-4}\, g$ of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius $10$ nm, then the number of gold particles per $mm^3$ of the sol will be

• A. $1.9 \times 10^{12}$
• B. $6 .3 \times 10^{14}$
• C. $6.3 \times 10^{10}$
• D. $2.4 \times 10^6$

Answer 1

Answer: (D)

Volume of the gold dispersed in one litre water
$=\frac{\text{Mass}}{\text{Density}} = \frac{1.9\times 10^{-4} \,g}{19\,g\,cm^{-3}}$
$ = 1 \times 10^{-5} \,cm^3$
Radius of gold sol particle $= 10 \times 10^{-9} m $
$= 10 \times 10^{-7} = 10^{-6} \,cm$
Volume of the gold sol particle $ = \frac{4}{3} \pi r^3$
$ = \frac{4}{3} \times \frac{22}{7} \times (10^{-6})^3 $
$ = 4.19 \times 10^{-18} \,cm^3$
No. of gold sol particle in $1 \times 10^{-5} cm^3$
$ = \frac{1 \times 10^{-5}}{4.19 \times 10^{-18}} = 2.38 \times 10^{12}$
No. of gold sol particle in one $mm^3$.
$ = \frac{2.38 \times 10^{12}}{10^6} = 2.38 \times 10^6$