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5 July, 05:22

How many grams of KCl are needed to prepare 0.750 L of a 1.50 M solution in water?

a. 83.9

b. 0.500

c. 1.13

d. 2.00

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Answers (1)
  1. 5 July, 05:27
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    Answer: Option A) 83.9g

    Explanation:

    KCl is the chemical formula of potassium chloride.

    Given that,

    Amount of moles of KCl (n) = ?

    Volume of KCl solution (v) = 0.75L

    Concentration of KCl solution (c) = 1.5M

    Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

    c = n / v

    make n the subject formula

    n = c x v

    n = 1.5M x 0.75L

    n = 1.125 mole

    Now given that,

    Amount of moles of KCl (n) = 1.125

    Mass of KCl in grams = ?

    For molar mass of KCl, use the molar masses of:

    Potassium, K = 39g;

    Chlorine, Cl = 35.5g

    KCl = (39g + 35.5g)

    = 74.5g/mol

    Since, amount of moles = mass in grams / molar mass

    1.125 mole = m / 74.5g/mol

    m = 1.125 mole x 74.5g/mol

    m = 83.81g

    Thus, 83.9 grams of KCl are needed to prepare 0.750 L of a 1.50 M solution in water
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