Question

A 15.0 mL sample of a Potassium Nitrate solution that has a mass of 15.78 g is placed in an evaporating dish and evaporated to dryness. The Potassium Nitrate that remains has a mass of 3.26 g. Calculate the following concentrations for the Potassium Nitrate solution
a. %(m/m),
b. % (m/v),
c. molarity (M)

Answer 1

Explanation:

A. % (m/m)

Mass of KNO3 solution = mass of water + mass of dry KNO3

%(m/m) = (mass of KNO3/total mass of KNO3 solution) * 100

= (3.26/15.78) * 100

= 20.66 %

B. %(m/v)

%(m/v) = (mass of dry KNO3/volume of KNO3 solution) * 100

= 3.26/15 * 100

= 21.73 %

C. molarity

Molar mass of KNO3 = 39 + 14 + (3*16)

= 101 g/mol

Number of moles = mass/molar mass

= 3.26/101

= 0.0323 mol.

Molarity is defined as the number of moles of the solute per unit volume of solution.

Molarity = 0.0323/0.015

= 2.15 M.

Answer 2

Answer:

a) 20.65%

b) 21.73%

c) 2.15 M

Explanation:

The mass conservation law in preparation or dilution of solutions, explains that the number of moles/mass of solute does not change after adding more solvent to dilute or removing more solvent to concentrate the solution.

The total volume of solution = 15.0 mL = 0.015 L

The total weight of solution = 15.78 g

The weight of the solute = 3.26 g.

The number of moles of KNO₃ = mass/molar mass of KNO₃

Molar mass of KNO₃ = 101.10 g/mol

Number of moles = 3.26/101.1 = 0.0322 mole

a) The weight percent in (%m/m) = 3.26/15.78 = 0.2065 = 20.65%.

b) The weight percent in (%mass/volume) = 3.26/15.0 = 0.2173 = 21.73%.

c) Molarity (mol/volume in L) = 0.0322/0.015= 2.15 M