Answer:
1. 0.45 mole
2. 49.95g
Explanation:
The following were obtained from the question:
Volume of solution = 300mL = 300/1000 = 0.3L
Molarity = 1.5 M
Mole of CaCl2 =?
1. We can obtain the mole of the solute as follow:
Molarity = mole of solute /Volume of solution
1.5 = mole of solute/0.3
Mole of solute = 1.5 x 0.3
Mole of solute = 0.45 mole
2. The grams in 0.45 mole of CaCl2 can be obtained as follow:
Molar Mass of CaCl2 = 40 + (35.5 x 2) = 40 + 71 = 111g/mol
Mole of CaCl2 = 0.45 mole
Mass of CaCl2 =?
Mass = number of mole x molar Mass
Mass of CaCl2 = 0.45 x 111
Mass of CaCl2 = 49.95g
Answer:
We have 0.45 moles CaCl2 in this 1.5 M solution
This 49.9 grams of CaCl2
Explanation:
Step 1: data given
Volume of the CaCl2 solution = 300 mL = 0.300 L
Molarity of the CaCl2 solution = 1.5 M
Molar mass CaCl2 = 110.98 g/mol
Step 2: Calculate number of moles in the solution
Moles CaCl2 = molarity solution * volume of solution
Moles CaCl2 = 1.5 M * 0.300 L
Moles CaCl2 = 0.45 moles
Step 3: Calculate mass CaCl2
Mass CaCl2 = moles CaCl2 * molar mass CaCl2
Mass CaCl2 = 0.45 moles * 110.98 g/mol
Mass CaCl2 = 49.9 grams CaCl2
We have 0.45 moles CaCl2 in this 1.5 M solution
This 49.9 grams of CaCl2
