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5 April, 23:14

A solution of salt (molar mass 90 g mol-1) in water has a density of 1.29 g/mL. The concentration of the salt is 35% by mass.

a. Calculate the molality of the solution.

b. Calculate the molarity of the solution.

c. Calculate the mole fraction of the salt in the solution.

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Answers (1)
  1. 6 April, 01:35
    0
    a) 5.9846 mol/kg

    b) 5.018 mol/L

    c) 0.09725

    Explanation:

    consider 100 g of solution

    now

    since the salt is 35 %, water will be 65 %

    now

    mass of salt = 35 g

    mass of water = 65 g

    we know that

    moles = mass / molar mass

    so

    moles of salt = 35 g / 90 (g/mol) = 0.389 mol

    moles of water = 65 / 18 = 3.6111 mol

    now

    volume of solution = mass of solution / density of solution

    volume of solution = 100 g / (1.29 g/ml)

    volume of solution = 77.52 ml

    volume of solution = 0.07752 L

    a) molality = moles of salt / mass of water (kg)

    molality = 0.389 mol / 0.065 kg

    molality = 5.9846 mol/kg

    b)

    molarity = moles of salt / volume of solution (L)

    molarity = 0.389 mol / 0.07752 L

    molarity = 5.018 mol/L

    c)

    now

    total moles in the solution = moles of salt + moles of water

    total moles in the solution = 0.389 + 3.6111

    total moles in the solution = 4 mol

    now

    mole fraction of salt = moles of salt / total moles

    mole fraction of salt = 0.389 mol / 4 mol

    mole fraction of salt = 0.09725
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