Question

The number
of moles of So4^2^- ions present in
250 ml
2M
aluminium
sulphate solution is​

Answer 1

Answer:

molarity=no of moles/volume of solution (in L).

2=(no of moles/ 250 ml)×1000

2=no of moles×4

no of moles=2/4=1/2= 0.5 mol

this is total moles of al2(so4)3.

now by using, concept of atomicity,

no of atoms of sulphate=no.of molecules×atomicity of sulphate

1 mole= 6.02×10^23 molecules

for,0.5 moles=(6.02/2)×10^23=3.01×10^23

atomicity of sulphate (2al+3(so4))

therefore, 3 is the atomicity of sulphate

now, no of sulphate atoms= 3×3.01×10^23

=9.03×10^23 atoms

now convert it into moles,

1 mole= 6.023×10^23 atoms ...............(1)

now ,[(1) /6.023×10^23]×9.03×10^23 on both sides,

(1×9.023×10^23)/6.023×10^23= 9.03×10^23

therefore,no of moles of sulphate= 1.5 moles