Question

The density of gold is 19 g/cm³. If 1.9 × 10^{−4 g} of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius 10 nm, the number of gold particles per mm³ of the sol will be ______ × 10⁶.

पर्याय

• 1.63

• 2.40

• 3.67

• 4.98

Answer 1

The density of gold is 19 g/cm³. If 1.9 × 10^{−4 g} of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius 10 nm, the number of gold particles per mm³ of the sol will be 2.40 × 10⁶.

Explanation:

Vol. of gold = `(1.9 xx 10^-4  "g")/(19  "g cm"^(-3))` = 10^–5 cm³

Vol. of gold = 10^–8 cm³,

Vol. of gold = 10^–17 m³

Now,

`"n"[4/3 xx π xx (10 xx 10^-9)^3]` = 10^–17

n = 2.4 × 10⁶