1. What mass of the following chemicals is needed to make the solutions indicated?

a. 1.0 liter of a 1.0 M mercury (II) chloride (HgCl2) solution

b. 2.0 liters of a 1.5 M sodium nitrate (NaNO3) solution

c. 5.0 liters of a 0.1 M Ca(OH)2 solution

d. 100 mL of a 0.5 M (NH4)3PO4 solution

2. Calculate the molarity of the following solutions.

a. 12 g of lithium hydroxide (LiOH) in 1.0 L of solution

b. 198 g of barium bromide (BaBr2) in 2.0 L of solution

c. 54 g of calcium sulfide (CaS) in 3.0 L of solution

3. Calculate the volume of each solution, in liters.

a. a 1.0 M solution containing 85 g of silver nitrate (AgNO3)

b. a 0.5 M solution containing 250 g of manganese (II) chloride (MnCl2)

c. a 0.4 M solution containing 290 g of aluminum nitrate (Al(NO3)3)

The following questions are answered below

Explanation:

1.a. 1.0 liter of a 1.0 M mercury (II) chloride (HgCl2) solution

  1.0 L 1 mol 271.49 g

    L        1 L    1 mol

= 271.49 g

= HgCl₂ = 271.49 g HgCl₂

1.b. 2.0 liters of a 1.5 M sodium nitrate (NaNO3) solution

= 255.03 g

= 255 g NaNo₃

1.c. 5.0 liters of a 0.1 M Ca(OH)2 solution

= 37 g Ca(OH)2

1.d. 100 mL of a 0.5 M (NH4)3PO4 solution

= 7.5 g (NH₄)₃PO₄

2. To find the molarity of the following solutions

2.a. 12 g of lithium hydroxide (LiOH) in 1.0 L of solution

= 0.50 m (LiOH)

2.b. 198 g of barium bromide (BaBr2) in 2.0 L of solution

= 0.33 m (BaBr₂)

2.c. 54 g of calcium sulfide (CaS) in 3.0 L of solution

= 0.25 m (CaS)

3. To find the volume of each solution

3.a. 1.0 M solution containing 85 g of silver nitrate (AgNO3)

= 0.50 L (AgNO₃)

3.b. 0.5 M solution containing 250 g of manganese (II) chloride (MnCl2)

= 4.0 L MnCl2

3.c. 0.4 M solution containing 290 g of aluminum nitrate (Al(NO3)3)

= 3.4 L (Al(NO₃)₃)

the correct answer is:

a. formation of precipitation