Question

1. You wish to prepare 184 grams of 3.45% NH4Cl.

U will need ___ grams of ammonium chloride and __mL of water.

* density of water is 1.00 g/mL

2. How many grams of KCH3COO are there in 201 grams of an aqueous solution that is 12.6% by weight KCH3COO

_______g KCH3COO

3. If 24.8 grand of an aqueous solution of Iron(III) sulfate, Fe2(SO4)3, contains 7.08 grams of iron(III) sulfate, what is the percentage by mass of iron(III) sulfate in the solution?

______% Fe2(SO4)3

4. An aqueous solution is made by dissolving 16.0 grams of zinc nitrate in 314 grams of water.

The molality of zinc nitrate in the solution is ____ m.

5. In the laboratory you are asked to make a 0.267 m iron(III) nitrate solution using 375 grams of water. How many grams of iron(III) nitrate should you add? ______ g

6. In the laboratory you are asked to make a 0.350 m nickel(II) acetate solution using 19.9 grams of nickel(II) acetate. How much water should you add? ______ g

Answer 1

1) mass of the solution = 184 g

Let mass of NH₄Cl = x g

Thus, % NH₄Cl = (mass of NH₄Cl/mass of solution)*100

or, mass of NH₄Cl = 0.0345*184 = 6.348 g

2) mass of KH₃COO = (12.6/100)*mass of solution = 0.126*201 = 25.326 g

3) % of iron(III) sulfate = (mass of iron(III)sulfate/mass of solution)*100 = (7.08/24.8)*100 = 28.55%

4) molality = moles of zinc nitrate/mass of water in kg

Now, molar mass of zinc nitrate = 127.38 g/mole

moles of zinc nitrate = mass/molar mass = 16/127.38 = 0.1256

Hence, molality = (0.1256/0.314) = 0.4 m

5) moles of iron (III) sulfate = molality*mass of water in kg = 0.267*0.375 = 0.1

molar mass of iron(III)sulfate = 400 g/mole

Thus, mass of iron (III) sulfate required = moles*molar mass = 0.1*400 = 40 g

6) molar mass of nickel(II)acetate = 176.7

Thus, moles of nickel(II)acetate = mass/molar mass = 19.9/176.7 = 0.113

Hence mass of water required = moles of nicekl(II)acetate/molality = 0.113/0.35 = 0.322 kg = 322 g