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10 March, 06:00

How many moles of solute are in 300 mL of 1.5 M CaCl2? How many grams fo CaCl2 is this?

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Answers (2)
  1. 10 March, 06:11
    0
    We have 0.45 moles CaCl2 in this 1.5 M solution

    This 49.9 grams of CaCl2

    Explanation:

    Step 1: data given

    Volume of the CaCl2 solution = 300 mL = 0.300 L

    Molarity of the CaCl2 solution = 1.5 M

    Molar mass CaCl2 = 110.98 g/mol

    Step 2: Calculate number of moles in the solution

    Moles CaCl2 = molarity solution * volume of solution

    Moles CaCl2 = 1.5 M * 0.300 L

    Moles CaCl2 = 0.45 moles

    Step 3: Calculate mass CaCl2

    Mass CaCl2 = moles CaCl2 * molar mass CaCl2

    Mass CaCl2 = 0.45 moles * 110.98 g/mol

    Mass CaCl2 = 49.9 grams CaCl2

    We have 0.45 moles CaCl2 in this 1.5 M solution

    This 49.9 grams of CaCl2
  2. 10 March, 09:57
    0
    1. 0.45 mole

    2. 49.95g

    Explanation:

    The following were obtained from the question:

    Volume of solution = 300mL = 300/1000 = 0.3L

    Molarity = 1.5 M

    Mole of CaCl2 = ?

    1. We can obtain the mole of the solute as follow:

    Molarity = mole of solute / Volume of solution

    1.5 = mole of solute/0.3

    Mole of solute = 1.5 x 0.3

    Mole of solute = 0.45 mole

    2. The grams in 0.45 mole of CaCl2 can be obtained as follow:

    Molar Mass of CaCl2 = 40 + (35.5 x 2) = 40 + 71 = 111g/mol

    Mole of CaCl2 = 0.45 mole

    Mass of CaCl2 = ?

    Mass = number of mole x molar Mass

    Mass of CaCl2 = 0.45 x 111

    Mass of CaCl2 = 49.95g
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