Question

Solution, dilution, concentration problems: Please answer 1-5 and show all work

1. How many grams of MgCl₂ are required to prepare 200 ml of a 0.35 M solution (MW of MgCl₂ = 95.2)?

2. Express the concentration of the solution in problem #1 in terms of g/liter.

3. Express the concentration of the solution in problem #1 in terms of %w/v.

4. A solution contains 160.0 g of ferrous ammonium sulfate (MW = 392.1) in 500 ml of solution. What is the molarity of the solution?

5. What is the percent by weight (%w/v) of the solution in #4?

Answer 1

1)

M = grams of MgCL2 / M.W. of MgCL2 x volume (in L)

Grams of MgCL2 = M x M.W. x volume

= 0.35 x 95.2 x 0.2

= 6.66 grams of MgCL2 is required.

2)

6.66 grams in 200 ml

so in 1000 ml = 1000 x 6.66 / 200 = 33.3 grams / L

3)

33.3 grams / L

so 3.33 grams per 100 mL

that is 3.33 % w/v

4)

M = grams / M.W. x volume

= 160 / 392.1 x 0.5

= 0.816 M

5)

160 grams in 500 mL

so 160/5

= 32 g in 100 mL

that is 32 %w/v