Question

10. You have 700 grams of Ni3(PO3)2. How many particles are present?

Answer 1

1) List the quantities.

Sample: 700 g Ni3(PO3)2

2) Convert gramas of Ni3(PO3)2 to moles of Ni3(PO3)2.

The molar mass of Ni3(PO3)2 is 334.0241 g/mol.

`mol\text{ }Ni_3(PO_3)_2=700\text{ }g\text{ }Ni_3(PO_3)_2*\frac{1\text{ }mol\text{ }Ni_3(PO_3)_2}{334.0241\text{ }g\text{ }Ni_3(PO_3)_2}=2.10\text{ }mol\text{ }Ni_3(PO_3)_2`

3) Convert moles to particles.

1 mol = 6.022*10^23 particles.

`parciles\text{ }Ni_3(PO_3)_2=2.10\text{ }mol\text{ }Ni_3(PO_3)_2*\frac{6.022*10^(23)\text{ }particles\text{ }Ni_3(PO_3)_2}{1\text{ }mol\text{ }Ni_3(PO_3)_2}=1.26*10^(24)\text{ }particles`

700 g Ni3(PO3)2 is equal to 1.26*10^24 particles.