Question

calculate the no of aluminium ion in 0.056 g of alumina

Answer 1

formula of alumina is Al2O3

molecule mass of alumina = 2×27 + 3×16 = 54+ 48 = 102 g

mole of alumina = 0.056 / 102 = 0.0005 mole

mole of aluminum = 0.0005 mole × atomicity

so, mole of aluminum = 0.00054 × 2 = 0.001 M

total no of Al ion = 0.001 × 6.02 × 10²³ × 3

total no of Al ion = 18.06 ×10^20 ion

Answer 2

Alumina is denoted by Al2O3

Molecular weight of al203 is 2x 27+16×3=54+48=102

We know that no of moles =weight given/molecular or atomic mass

Therfore,no of moles=0.056/102

We get no of moles =0.0005 which is really not very pleasing

We also know 1 mole of al203 contains 6.022×10^23

Therefore,0.0005moles wilk contain 0.0005x6.022x10^23

But al is present as 2 in the molecule of al203 therefore no of moles

Of al in alumina is 0.0005 x 2 Na molecules

Therfore ,no of al3+ions in alumina is 0.001 x 6.023x10^23 x3 i.e

6.023×10^20 x 3= 18.069 x 10^20(since there r three positive cations to be considered)