Question

The density of gold is $19g/{cm}^3$. 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

The correct option is D $2.4\times {10}^6$

Volume of gold present in solution,

$=\frac{Massofgold}{Densityofgold}=\frac{1.9\times {10}^{-4}g}{19g/{cm}^3}=0.1\times {10}^{-4}{cm}^3$

For spherical particles of gold with a radius equal to $10nm$

The volume of each particle:
$\frac{4}{3}\pi r^3=\frac{4}{3}\times \frac{22}{7}\times {(10\times {10}^{-7}cm)}^3$

$=\frac{88}{21}\times {10}^{-18}{cm}^3$

Number of gold particles present:

$=\frac{Volumeofgoldinsolution}{Volumeofeachparticle}$

$=\frac{0.1\times {10}^{-4}{cm}^3}{\frac{88}{21}\times {10}^{-18}{cm}^3}$

$=\frac{21}{88}\times {10}^{13}=2.4\times {10}^{12}$

$2.4\times {10}^{12}$ particles of gold are presnt in $1000{cm}^3$ (1 litre).

$\therefore$ Number of particles present per ${mm}^3$

$=\frac{2.4\times {10}^{12}}{{10}^6}[1L={10}^6{mm}^3]$

$=2.4\times {10}^6$

Hence, the correct answer is option $\text{D}$.