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27 July, 07:47

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)

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Answers (2)
  1. 27 July, 07:59
    0
    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.
  2. 27 July, 10:51
    0
    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
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