Question

calculate the no. of ca, c and o atoms present in 10g of caco3​

Answer 1

So, in 10g of calcium carbonate, the molecular mass of CaCO3

= ( 40 + 12 + 48 ) g/mol = 100 g/mol

No. of moles = W in g / W in g/mol

= 10 g / 100 g/mol

= 0.1 mol

1 mole of Oxygen = 6.022 × 10²³

0.1 mole of O = 0.1 × 6.022 × 10²³

0.1 mole of 3O ( because it is CaCO3 )

= 6.022 ×10²² × 3

= 18.066 × 10²² or

= 1.8066 × 10¹ × 10²²

=> 1.81 × 10²³ atoms

Now, as it is siad that

No. of molecules of O2 = No. of molecules of N2

W in g/ 32 × Avagrado no. = 1.4 g / 28 g/mol × Avagrado no.

W in g = 1.4 / 28 ×32

= 1.6 g

=> Hence, 1.6 g of oxygen molecules will contain same no.of molecules as in 1.4g of nitrogen.

Have great future ahead!