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 solution 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 $2.4\times {10}^6$

It is given that the density of gold is $19$ g/cm${}^3$.

$1.9\times {10}^{-4}$ g of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius $10$ nm.

Volume of gold used $=\frac{m}{d}=\frac{1.9\times {10}^{-4}}{19}=1\times {10}^{-5}$ cm${}^3$

Volume of one gold particle $=\frac{4}{3}\pi r^3$ $=\frac{4}{3}\times \frac{22}{7}\times (10\times {10}^{-7}{)}^3$ $=4.19\times {10}^{-18}$ cm${}^3$

$\therefore$ Total number of particles $=\frac{1\times {10}^{-5}}{4.19\times {10}^{-18}}=2.4\times {10}^{12}$

Thus, $2.4\times {10}^{12}$ particles of gold are present in $1$ litre or ${10}^{-3}$ m${}^3$ of volume.

$\therefore$ The number of particles per mm${}^3=\frac{2.4\times {10}^{12}}{{10}^{-3}\times {10}^9}$ $=2.4\times {10}^6$

Hence, option D is the correct answer.