Appendix A: Questions L A0.25 Maqueous solution of Calz is Cars and M in 2. A0.7

Question

Appendix A: Questions L A0.25 Maqueous solution of Calz is Cars and M in 2. A0.75 M aqueous solution of Fe(CIO), is Min Fe and in C0 3. A25M aqueous solution of Fe(S0a, is M in Fe andn SO An aqueous solution of Al(S0 Al[S0.)s in this solution. is 036 M in S0, Calculate the concentration (in M) of . Calculate the concentrations of K and NOs in an aqaeous solution prepared by dissolving 30.0 g KNOs in enough water to make 300 mL of solution 6. Caloulate the concentrations of A and SO in an aqueous solution prepared by dissolving 17.1 g Aldso.hin enough water to make 400 mL of solution. . Calcalate the concentrations of Nar and SO in an aqueous solution prepared by dissolving 852 g NauS0, in enough water to maloe 4.00 L of solution.

Explanation / Answer

1)

CaI2 ----> Ca+2 + 2I-

Hence in 1 liter of the solution if one mole of CaI2 is added it effectively contains one mole of Ca+2 ions and two moles of I- ions.

Hence answer is 0.25M, 0.5M.

2)

Fe(ClO4)3 ----> Fe+3 + 3ClO4-

Hence in 1 liter of the solution if one mole of Fe(ClO4)3 is added it effectively contains one mole of Fe+3 ions and three moles of ClO4- ions.

Hence answer is 0.75M, 2.25M.

3)

Fe2(SO4)3 ----> 2Fe+3 + 3SO4-2

Hence in 1 liter of the solution if one mole of Fe2(SO4)3 is added it effectively contains two mole of Fe+3 ions and three moles of SO4-2 ions.

Hence answer is 5M, 7.5M.

4)

Al2(SO4)3 ----> 2Al+3 + 3SO4-2

Hence in 1 liter of the solution if one mole of Al2(SO4)3 is added it effectively contains two mole of Al+3 ions and three moles of SO4-2 ions.

Therefore for 0.36M sulphate ion concentration there has to be 0.36/3 = 0.12M of Al2(SO4)3

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