Question

Convert 0.12M of cobalt (II) nitrate + 0.0386M of chromium (III) Nitrate solution to percentage concentration of each component

Chromium (400.14g/mol)

Cobalt (291.2g/mol)

Answer 1

there are two types of % concentration I can give a) concentration in terms of Mole % ,b) concentration in terms of Mass %

a) concentration in terms of Mole % = [moles of A/ total moles(A+B) ] * 100

total moles = 0.12 + 0.0386 = 0.1586 M

So Mole % of cobalt (II) nitrate = (0.12 / 0.1586) * 100 = 75.66 %

Mole % chromium (III) Nitrate = (0.0386 / 0.1586) * 100 = 24.34 %

a) concentration in terms of Mass % = [Mass of A/ total mass(A+B) ] * 100

mass of cobalt (II) nitrate = 0.12 * 291.2 = 34.944 g

mass of chromium (III) Nitrate = 0.0386 * 400.14 = 15.445 g

Total mass = 34.944 + 15.445 g =50.39g

So Mass % of cobalt (II) nitrate = (34.944 / 50.39) * 100 = 69.35 %

Mole % chromium (III) Nitrate = (15.445 / 50.39) * 100 = 30.65 %

[Note: check your given M.wt

Chromium (400.14g/mol) Cobalt (291.2g/mol) is not the actual values is it given by your teacher or you made mistake, if it is by yours redo the calculation the same way i did)