Question

calculate the number of atoms present in 6.2 g of phosphorous. ​

Answer 1

$\sf \huge{\fbox{SOLUTION :}}$

We know that ,

$\sf \large \fbox{Number \: of \: atoms = \frac{Given \: mass}{atomic \: mass} \times avogadro \: number}$

Substitute the known values , we get

$\sf \mapsto Number \: of \: atoms = \frac{6.2}{31} \times 6.022 \times {(10)}^{23} \\ \\ \sf \mapsto Number \: of \: atoms = \frac{62}{310} \times 6.022 \times {(10)}^{23} \\ \\ \sf \mapsto Number \: of \: atoms = 2 \times 6.022 \times {(10)}^{22} \\ \\ \sf \mapsto Number \: of \: atoms = 6.022 \times {(20)}^{22} \: \: moles$

Hence , 6.022 × (20)²² moles present in 6.2 g of phosphorous

Answer 2

Answer:

Explanation:

No. Of atoms = the given mass/the atomic number *Avogadro number

Number of atoms =6.2/31*6.022*10 ^23

No. Of atoms is 12.044 *10^22