Question

How many atoms are present in 114 gm of Al2(SO4)3 ?

Answer 1

Molar mass of compound= 342g
Total atoms in compound= 2+3+12= 17

Number of moles of compound
= given mass/ molar mass
= 114/342
= 1/3 mol

Hence, number of moles of atoms
= 17 x 1/3

In terms of Avogadro’s number
= 17 x 1/3 x 6.02 x 10^23
= 3.71 x 10^24 atoms.

Hope it helps!

Answer 2

Given:

The mass of Al2(SO4)3 = 114 gm

To Find:

The no of atoms in the given mass of Al2(SO4)3.

Calculation:

- The molar mass of Al2(SO4)3 = 342 gm

- The no of moles of Al2(SO4)3 = 114/342 = 0.33 moles

- The no of atoms in 1 molecule = 2 + 5×3 = 17

- The no of atoms in 0.33 moles of Al2(SO4)3 = 0.33 × 17 × 6.022 × 10²³

⇒ n = 33.783 × 10²³

⇒ n = 3.3783 × 10²⁴

- So, the no of atoms present in 114 gm of Al2(SO4)3 are 3.3783 × 10²⁴.