Question

The density of gold is $19g/c{m}^3.$ If $1.9\times {10}^{-4}g$ of gold is dispersed in $1L$ of water to give a sol having spherical gold particles of radius $10nm$, the number of gold particles per $m{m}^3$ of the sol will be ___ $\times {10}^6$.

Answer 1

$\text{Volume of the gold dispersed in}1L\text{water}=\frac{m}{d}=\frac{1.9\times {10}^{-4}}{19}=1\times {10}^{-5}c{m}^3$
$\text{Radius of gold particle}=10nm={10}^{-6}cm$
$\therefore \text{Volume}=\frac{4}{3}\pi r^3=\frac{4}{3}\times \frac{22}{7}\times ({10}^{-6}{)}^3=4.19\times {10}^{-18}c{m}^3$
$\begin{array}{cc}\text{Number of particles in}1\times {10}^{-5}c{m}^3& =\frac{1\times {10}^{-5}}{4.19\times {10}^{-18}}\\ & =2.38\times {10}^{12}\end{array}$
$\begin{array}{cc}\text{Number of particles in}1m{m}^3& =\frac{2.38\times {10}^{12}}{{10}^6}\\ & =2.38\times {10}^6\end{array}$