Question

Find the number of electrons in a drop of sulphuric acid weighing

4.9 ×10-3 mg.

​

Answer 1

The number of electrons = 1.50575 × $10^{18}$ electrons

Explanation:

Given,

The mass of one drop of $H_2SO_4$ = 4.9 × $10^{-3}$ mg = 4.9 × $10^{-6}$ g and

The moecular mass of $H_2SO_4$ = 2(1) + 32 + 16(4) = 98

To find, the number of electrons = ?

The number of moles of $H_2SO_4$  in a drop = 4.9 × $10^{-6}$ g/ 98

= 0.05 × $10^{-6}$ moles

The total number of electrons in $H_2SO_4$ ​=  50

In  1 mole of  number of electrons = 50 × Avogadro's Number

= 50 × 6.022 × $10^{23}$ electrons

In 1 drop, number of electrons = 50 × 6.022 × $10^{23}$ × 0.05 × $10^{-6}$

= 1.50575 × $10^{18}$ electrons