Question

How many moles of solute are in 300 mL of 1.5 M CaCl2? How many grams fo CaCl2 is this?

Answer 1

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

Answer 2

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