Question

. A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl2 in this solution is __________ molar.
a. 0.564
b. 0.571
c. 0.569
d. 0.537
e. 0.214

Answer 1

Answer:

The concentration of CaCl2 in this solution is 0.564 molar (option A)

Explanation:

Step 1: Data given

Mass of CaCl2 = 23.7 grams

Mass of water = 375 grams

Density of solution is 1.05 g/mL

Step 2: Calculate total mass

Total mass = mass of CaCl2 + mass of water

Total mass = 23.7 grams + 375 grams = 398.7 grams

Step 3: Calculate volume of the solution

Density = mass / volume

Volume = mass / density

Volume = 398.7 grams / 1.05 g/mL

Volume = 379.7 mL = 0.3797 L

Step 4: Calculate moles CaCl2

Moles CaCl2 = mass CaCl2 / molar mass CaCl2

Moles CaCl2 = 23.7 grams / 110.98 g/mol

Moles CaCl2 = 0.214 moles

Step 5: Calculate concentration

Concentration of CaCl2 = moles / volume

Concentration of CaCl2 = 0.214 moles / 0.3797 L

Concentration of CaCl2 = 0.564 mol / L = 0.564 molar

The concentration of CaCl2 in this solution is 0.564 molar (option A)