The density of gold is 19 g/cm^3. If 1.9 tim | vedclass

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(D) Volume of gold present in solution = $\frac{	ext{Mass of gold}}{	ext{Density of gold}} = \frac{1.9 	imes 10^{-4} \ g}{19 \ g/cm^3} = 1.0 	imes

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$2.4 	imes 10^6$

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(D) Volume of gold present in solution = $\frac{	ext{Mass of gold}}{	ext{Density of gold}} = \frac{1.9 	imes 10^{-4} \ g}{19 \ g/cm^3} = 1.0 	imes 10^{-5} \ cm^3$.For spherical particles of gold with radius $r = 10 \ nm = 10 	imes 10^{-7} \ cm = 10^{-6} \ cm$,the volume of each particle = $\frac{4}{3} \pi r^3 = \frac{4}{3} 	imes 3.14 	imes (10^{-6} \ cm)^3 = 4.186 	imes 10^{-18} \ cm^3$.Number of gold particles present in $1 \ L$ $(1000 \ cm^3)$ = $\frac{	ext{Total volume of gold}}{	ext{Volume of one particle}} = \frac{1.0 	imes 10^{-5} \ cm^3}{4.186 	imes 10^{-18} \ cm^3} \approx 2.389 	imes 10^{12} 	ext{ particles}$.Since $1 \ L = 1000 \ cm^3 = 10^6 \ mm^3$,the number of particles per $mm^3$ = $\frac{2.389 	imes 10^{12}}{10^6} \approx 2.4 	imes 10^6 	ext{ particles/mm}^3$.

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

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