Question

6. A 25.0-mL sample of potassium chloride solution was found to have a mass of 25.225 g.

After the solution was evaporated to dryness, the mass of the KCl residue was 1.396 g.
Calculate the (a) mass/mass percent concentration and (b) molarity of the solution.
(a)
(b)
7. (optional) A hospital saline solution is 0.90% NaCl. What is the molar concentration of
0.00% caline (58 44 g/mol) solution? (Assume the density of the solution is 1.01 g/mL.)

Answer 1

6. a. 5.53 (m/m) %; b. 0.7490M

7. 0.156M

Step-by-step explanation:

6. In the solution of KCl, there are 1.396g of KCl in 25.225g of solution. As mass/mass percent is:

Mass solute / Mass solution * 100

The mass/mass percent of KCl is:

a. 1.396g KCl / 25.225g solution * 100

5.53 (m/m) %

b. Molarity is moles of solute per liters of solution:

Moles KCl -Molar mass: 74.55g/mol-:

1.396g KCl * (1mol / 74.55g) = 0.018726 moles KCl

Volume in liters: 25mL = 0.025L

Molarity:

0.018726 moles KCl / 0.025L = 0.7490M

7. 0.90% means 0.90g of NaCl in 100g of solution:

Moles NaCl -Molar mass: 58.44g/mol-:

0.90g NaCl * (1mol / 58.44g) = 0.0154 moles NaCl

Volume in Liters:

100g * (1mL / 1.01g) = 99mL = 0.099L

Molarity:

0.0154 moles NaCl / 0.099L =

0.156M