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

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(D) Volume of the gold dispersed in $1 \ L$ of water $= \frac{	ext{Mass}}{	ext{Density}} = \frac{1.9 	imes 10^{-4} \ g}{19 \ g/cm^3} = 1 	imes 10^

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$2.4 	imes 10^6$

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(D) Volume of the gold dispersed in $1 \ L$ of water $= \frac{	ext{Mass}}{	ext{Density}} = \frac{1.9 	imes 10^{-4} \ g}{19 \ g/cm^3} = 1 	imes 10^{-5} \ cm^3$. Radius of gold sol particle $= 10 \ nm = 10 	imes 10^{-7} \ cm = 10^{-6} \ cm$. Volume of one gold sol particle $= \frac{4}{3} \pi r^3 = \frac{4}{3} 	imes 3.1416 	imes (10^{-6} \ cm)^3 \approx 4.19 	imes 10^{-18} \ cm^3$. Number of gold sol particles in $1 \ L$ $(1000 \ cm^3)$ $= \frac{1 	imes 10^{-5} \ cm^3}{4.19 	imes 10^{-18} \ cm^3} \approx 2.38 	imes 10^{12}$. Since $1 \ L = 10^6 \ mm^3$,the number of particles per $mm^3 = \frac{2.38 	imes 10^{12}}{10^6} = 2.38 	imes 10^6 \approx 2.4 	imes 10^6$.

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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