Answer:
a) 7.40 grams
b) 0.213 M
c) 102 moles
Explanation:
(a) Mass (g) of solute in 175.4 mL of 0.267 M calcium acetate
Step 1: Data given
Volume = 175.4 mL = 0.1754 L
Molarity = 0.267 M
Molar mass = 158.17 g/mol
Step 2: Calculate moles
Moles = molarity * volume
Moles = 0.267 M * 0.1754 L
Moles = 0.0468 moles
Step 3: Calculate mass
Mass = 0.0468 moles * 158.17 g/mol
<u>Mass = 7.40 grams</u>
b) Molarity of 597 mL solution containing 21.1 g of potassium iodide
Step 1: Data given
Volume = 597 mL = 0.597 L
Mass = 21.1 grams
Molar mass KI = 166.0 g/mol
Step 2: Calculate moles KI
Moles KI = 21.1 grams / 166.0 g/mol
Moles KI = 0.127 moles
Step 3: Calculate molarity
Molarity = moles / volume
Molarity = 0.127 moles / 0.597 L
Molarity = 0.213 M
(c) Amount (mol) of solute in 145.6 L of 0.703 M sodium cyanide
Step 1: Data given
Volume = 145.6 L
Molarity = 0.703 M
Step 2: Calculate moles
moles = molarity * volume
Moles = 0.703 M * 145.6 L
Moles = 102 moles
