Surface ChemistryHard

Question

A quantity of 1.9 × 10−4 g of the metal having density 19 g/ml is dispersed in 1 L of water to give a sol having spherical metal particles of radius 10 nm. The approximate number of metal sol particles per cm3 of the sol is

Options

A.2.39 × 109
B.4 × 1010
C.1.9 × 109
D.2.8 × 10−12

Solution

Volume of metal used = $\frac{1.9 \times 10^{- 4}}{19} = 10^{- 5}\text{c}\text{m}^{3}$

$\therefore N \times \frac{4}{3} \times \left( 10 \times 10^{- 7}\text{cm} \right)^{3} = 10^{- 5}\text{c}\text{m}^{3} $$$\Rightarrow N = 2.39 \times 10^{12}$$

Hence, number of particles per cm3

$= \frac{2.39 \times 10^{12}}{1000} = 2.39 \times 10^{9}$

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