Question

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