Chemistry: Molarity

Question

Calculate the molarity of each of the following solutions:

(a) 293g HCl in 666mL of solution, a concentrated HCl solution

(b) 2.0226 g FeCl₃ in 0.1250L of a solution used as an unknown in general chemistry laboratories.

(c) 0.001mg Cd⁺² in 0.100L, the maximum permissible concentration of cadmium in drinking water

(d) 0.0079g C₇H₅SNO₃ in one ounce(29.6 mL), the concentration of Saccharin in the diet soft drink.

Answer 1

You MUST understand what molarity (M) is. It is simply defined as moles of solute per liter of solution. So, for each situation, you must find the MOLES of solute, the LITERS of SOLUTION, and then divide one by the other.

(a) 293g HCl in 666mL of solution, a concentrated HCl solution:

moles HCl = 293 g HCl x 1 mole HCl/36.46 g HCl = 8.036 moles HCl

liters of solution = 666 ml x 1 L/1000 ml = 0.666 L

Molarity = 8.036 moles/0.666 L = 12.1 M (to 3 significant figures)

(b) 2.0226 g FeCl₃ in 0.1250L of a solution:

moles FeCl3 = 2.0226 g FeCl₃ x 1 mole FeCl₃ / 162.2 g FeCl3 = 0.01247 moles FeCl₃

liters of solution = 0.1250 L

Molarity = 0.01247 moles / 0.1250 L = 0.09976 M (to 4 significant figures)

(c) 0.001mg Cd⁺² in 0.100L:

moles Cd²⁺ = 0.001 mg x 1 g/1000 mg x 1 mol Cd / 112.41 g = 8.896x10^-9 moles Cd²⁺

liters of solution = 0.100 L

Molarity = 8.896x10^-9 moles / 0.100 L = 8.896x10^-8 M = 8.90x10^-8 M (to 3 significant figures)

(d) 0.0079g C₇H₅SNO₃ in one ounce(29.6 mL):

moles C₇H₅SNO₃ = 0.0079 g x 1 mol/183.18 g = 4.313x10^-5 moles

liters of solution = 29.6 ml x 1 L/1000 ml = 0.0296 L

Molarity = 4.313x10^-5 moles / 0.0296 L = 1.5x10^-3 M (to 2 significant figures)