The density of gold is 19 g/cm3. If 1.9x10-4 g of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius l0 nm, then the number of gold particles per mm3 of the sol will be :

A
$1.9 \times 10^{12}$
B
$6.3 \times 10^{14}$
C
$6.3 \times 10^{10}$
D
$2.4 \times 10^{6}$

Solution:

Volume of gold dispersed in 1 Lwater
$= \frac{Mass}{Density} = \frac{1 .9 \times 10^{- 4} g}{19 gm {cm}^{- 3}} = 1 \times 10^{- 5} {cm}^{3}$
Radius of gold sol particle
$= 10 nm = 10 \times 10^{- 7} cm = 10^{- 6} cm$
Volume of gol sol particle = $\frac{4}{3} {πr}^{3}$
$= \frac{4}{3} \times \frac{22}{7} \times {(10^{- 6})}^{3} = 4 .19 \times 10^{- 18} {cm}^{3}$
$∴$ No. of gold sol particles in $1 \times 10^{- 5} {cm}^{3}$
$= \frac{1 \times 10^{- 5}}{4.19 \times 10^{- 18}} = 2.38 \times 10^{12}$
$∴$ No. of gold sol particles in one mm³
$= \frac{2.38 \times 10^{12}}{10^{6}} = 2.38 \times 10^{6}$

The density of gold is 19 g/cm³. If 1.9x10^-4 g of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius l0 nm, then the number of gold particles per mm³ of the sol will be :

Solution:

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