Calculate the number of moles and molecules in 19.86 g. of Pb(NO₃)₂. [Pb = 207, N = 14, O = 16]

Gram molecular mass of Pb(NO₃)₂
= Pb + 2 [N x 3(O)]
= Pb + 2N + 6O
= 207 + (2 x 14) + (6 x 16)
= 207 + 28 + 96 = 331

As,

331 g of Pb(NO₃)₂ = 1 mole
Therefore, 19.86 g

$= \dfrac{1}{331} \times 19.86 = 0.06 \text { moles} \\[0.5em]$

Hence, number of moles in 19.86 g. of Pb(NO₃)₂ = 0.06 moles

As,

1 mole of Pb(NO₃)₂ weighs 331 g and has 6.023 x 10²³ molecules

Therefore, 19.86 g of Pb(NO₃)₂ will have

$= \dfrac{ 6.023 \times 10^{23}}{331} \times 19.86 \\[0.5em] = \dfrac{19.86}{331} \times 6.023 \times 10^{23} \\[0.5em] = 0.06 \times 6.023 \times 10^{23} \text{molecules}$

Hence, number of molecules in 19.86 g. of Pb(NO₃)₂ = 0.06 x 6.023 x 10²³ moles