Question

Calculate the number of moles present in (take value of Avagdro number 6*10^23)
a. 1.2 *10^20 molecules of CO
b. 1.5*10^16 atoms of Ca​

Answer 1

$\bold{Formula} \\ \boxed{n = \frac{N}{N_{A}}} \\ where, \\ N = no. \: of \: particles \\ N_{A} = avagdro \: number$

$\bold{a. \: 1.2 \times {10}^{20} \: molecules \: of \: CO} \\ = > n = \frac{1.2 \times {10}^{20} }{6 \times {10}^{23} } \\ = > n = \frac{ \cancel12 \times {10}^{20} }{ \cancel60 \times {10}^{23} } \\ = > n = 0.2 \times {10}^{20} - {10}^{23} \\ = > n = 0.2 \times {10}^{ - 3} \\ = > \boxed{n = 2 \times {10}^{ - 4} }$

$\bold{b.1.5 \times {10}^{16} \: atoms \: of \: Ca}\\ = > n = \frac{1.5 \times {10}^{16} }{6 \times {10}^{23} } \\ = > n = \frac{ \cancel15 \times {10}^{16} }{ \cancel60 \times {10}^{23} } \\ = > n = \frac{1 \times {10}^{16} }{4 \times {10}^{23} } \\ = > n = 0.25 \times {10}^{16} - {10}^{23} \\ = > n = 0.25 \times {10}^{ - 7} \\ = > \boxed{n = 25 \times {10}^{ - 9}}$

Answer 2

Answer :-

To Find:- No. of mole present in the given molecule

Solution :- By using the formulae n = N/Na

Where n stands for No. of mole

N = No. of particles

Na = Avagdro Number

n = 1.2 × 10^20/ 6 × 10^23

n = 0.2 × 10^20 - 10^23

n = 0.2 × 10^-3

n = 2 × 10^-4

Therefore, the no. of mole present in 1.2 × 10^20 molecule of atom is 2 × 10^-4.

b. 1.5 × 10^16

Using same formulae We get,

n = 1.5 × 10^16/ 6 × 10^23

n = 1 × 10^16/ 4 × 10^23

n = 0.25 × 10^-7

n = 25 × 10^-9

Therefore, the no. of mole present in 1.5 × 10^16 atom of Ca (Calcium ) is 25 × 10^-9.