Question

What is the Concentration of [Cu⁺] in 1.43mL of 0.0098M Cu₂SO₄ to 100.00mL sample?

What is the concentration of [F^-] in 12.43mL sample og 650.2mL that is 0.0342M of FeF₃?

Find the moles of S^2- are in 16.00mL of 0.060M Na₂S?

Answer 1

Concentration of [Cu+] in 1.43mL of 0.0098M Cu2SO4 to 100.00mL

V1 X N1 = V2 X N2

V1 = 1.43 ml

N1 = 0.0098 M

V2 = 100 ml

N2 = ?

Substitute in the formula and find N2

N2 = 1.43 x 0.0098 / 100 = 0.00014014 M

But in each molecule of Cu2SO4 have 2 Cu+ ions are present

Concentration of [Cu+] in 100 ml = 0.00014014 M x 2 = 0.00028028 M

concentration of [F-] in 12.43mL sample og 650.2mL that is 0.0342M of FeF3

V1 X N1 = V2 X N2

V1 = 12.43 ml

N1 = 0.0342

V2 = 650.2 ml

N2 = ?

Substitute in the formula and find N2

N1 = 12.43 x 0.0342 / 650.2  = 6.5380 x 10-4 Mole x 3 = 0.0019614 M

because each FeF₃ sample has 3 F^- ions

concentration of [F-] in 12.43mL = 0.0019614 M

moles of S2- are in 16.00mL of 0.060M Na2S

moles of S^2- = 16 x 0.06 /1000 = 0.00096 M