Question

4.62 ) (a) Calculate the molarity of a solution made by dissolving 12.5 grams of Na2CrO4 in enough water to form exactly 750 mL of solution. (b) How many moles of KBr are present in 150 mL of a 0.112 M solution

Answer 1

Answer:

a) 0.103 M

b) 0.0168 moles KBr

Explanation:

Step 1: Data given

Mass of Na2CrO4 = 12.5 grams

Molar mass of Na2CrO4 = 161.97 g/mol

volume = 750 mL = 0.750 L

Step 2: Calculate moles Na2CrO4

Moles Na2CrO4 = Mass Na2CrO4 / molar mass Na2CrO4

Moles Na2CrO4 = 12.5 grams / 161.97 g/mol

Moles Na2CrO4 = 0.0772 moles

Step 3: Calculate molarity of the solution

Molarity = moles / volume

Molarity = 0.0772 moles / 0.750 L

Molarity = 0.103 M

(b) How many moles of KBr are present in 150 mL of a 0.112 M solution

Moles = molarity * volume

Moles KBr = 0.112 M * 0.150 L

Moles KBr = 0.0168 moles

Answer 2

Answer:

a. [Na₂CrO₄] = 0.10 M

b. 0.017 moles of KBr

Explanation:

Molarity means a sort of concentration which indicates the moles of solute over 1L of solution.

We determine the moles of solute: 12.5 g / 162g/mol = 0.0771 moles

We convert the volume of solution from mL to L = 750 mL . 1L/1000mL = 0.750L

Molarity (mol/L) → 0.0771 mol / 0.750L = 0.10 M

b. In order to determine the moles of solute, with the molarity of solution and the volume we assume:

Molarity = moles of solute /volume of solution

Then, Molarity . Volume of solution (L) = moles of solute

We convert the volume of solution from mL to L = 150 mL . 1L/1000mL = 0.150L

0.112 mol/L . 0.150L = Moles of solute → 0.017 moles of KBr