Question

calculate the no. of particles in each of the following. (i) 46g of sodium atom. (ii) 8g of oxygen. (iii) 0.1 mole of oxygen​

Answer 1

Answer :-

(i) 1.2044 × 10²⁴ particles .

(ii) 3.011 × 10²³ particles .

(iii) 6.022 × 10²² particles .

Explanation :-

For number (i) :-

Molar mass of Sodium is 23 g/mol.

So required number of particles in 46 g of it will be :-

= Number of moles × Avogadro Number

= (46/23) × 6.022 × 10²³

= 2 × 6.022 × 10²³

= 12.044 × 10²³

= 1.2044 × 10²⁴

For number (ii) :-

Molar mass of Oxygen is 16 g/mol and given mass is 8 grams. Thus, no of particles :-

= (8/16) × 6.022 × 10²³

= 0.5 × 6.022 × 10²³

= 3.011 × 10²³

For number (iii) :-

Here we are given with 1 mole of Oxygen . So we will put values again and obtain it as :-

= 0.1 × 6.022 × 10²³

= 0.6022 × 10²³

= 6.022 × 10²²

Answer 2

Answer:

(I)

• Mass of sodium atom = 46 g

Moles of Na atoms :

$\implies n_{Na} = \dfrac{46}{23} \\ \\ \implies n_{Na} = 2 \: mol$

No. of particles:

$\implies n_{Na} = \dfrac{ no. \: of \: particles}{6.022 \times {10}^{23} }$

$\implies no. \: of \: particles = 2 \times 6.022 \times {10}^{23}$

$\implies \bf no. \: of \: particles = 12.044 \times {10}^{23}$

(ii)

• Mass of oxygen = 8g

Moles of Oxygen:

$\implies n_{O} = \dfrac{8}{16} \\ \\ \implies n_{Na} = 0.5 \: mol$

No. of particles:

$\implies n_{O} = \dfrac{ no. \: of \: particles}{6.022 \times {10}^{23} }$

$\implies no. \: of \: particles = 0.5 \times 6.022 \times 10^{23} \\ \\ \implies \bf no. \: of \: particles = 3.011 \times 10^{23}$

(iii)

• moles of oxygen = 0.1 mol

No. of particles of oxygen:

$\implies no. \: of \: particles = 0.1 \times 6.022 \times 10^{23}$

$\implies \bf no. \: of \: particles = 0.6022 \times 10^{23}$