Question

Find the number of Nitrogen atoms present in 65 mg of NaN3​

Answer 1

Welcome Student,

◆ Answer -

1.8066×10^21 N-atoms

● Explanation -

# Given -

W = 65 mg = 0.065 g

# Solution -

Molar mass of NaN3 is given as -

M = 23 + 3 × 14

M = 65 g/mol

Moles of NaN3 can be calculated as -

n = W / M

n = 0.065 / 65

n = 10^-3 mol

1 mole of NaN3 contains 3 N-atoms, thus 10^-3 moles NaN3 consists of -

N = 3 × 10^-3 × NA

N = 3 × 10^-3 × 6.022×10^23

N = 1.8066×10^21 atoms

Hence, 65 mg of NaN3 contains 1.8066×10^21 N-atoms.

Thanks dear.

Answer 2

given, mass of NaN3 = 65mg = 65 × 10^-3 g [ as 1mg = 10^-3 g ]

molecular mass of NaN3 = atomic mass of Na + 3 × atomic mass of N

= 23g/mol + 3 × 14g/mol

= (23 + 42)g/mol

= 65g/mol

so, number of moles in 65mg of NaN3 = given mass of NaN3/molecular mass of NaN3

= 65 × 10^-3/65

= 10^-3 mol

in one mole of NaN3, three moles of nitrogen atoms are present.

so, in 10^-3 mol of NaN3 , number of moles of nitrogen atoms = 3 × 10^-3 mol

now, number of nitrogen atoms present in 65mg of NaN3 = number of moles of nitrogen atoms × Avogadro's number.

= 3 × 10^-3 × 6.023 × 10²³

= 1.8069 × 10²¹ .