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13 July, 02:24

For an aqueous solution of sodium chloride (NaCl), determine the molarity of 4.15 L of a solution that contains 173 g of sodium chloride. Determine the volume of this solution that would contain 3.93 moles of sodium chloride. Determine the number of moles of sodium chloride in 22.45 L of this solution.

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  1. 13 July, 02:52
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    a) Molarity = 0.713 M

    b) volume = 5.51 L

    c) Number of moles = 16.01 moles

    Explanation:

    A) What is the molarity of a 4.15L solution that contains 173 g of sodium chloride?

    Step 1: Data given

    Volume = 4.15 L

    Mass of NaCl = 173 grams

    Molar mass of NaCl = 58.44 g/mol

    Step 2: Calculate moles of NaCl

    Number of moles NaCl = Mass NaCl / molar mass

    Moles NaCl = 173 grams / 58.44 g/mol

    Moles NaCl = 2.96 moles

    Step 3: Calculate molarity of solution

    Molarity = moles NacL / volume

    Molarity = 2.96 moles / 4.15L

    Molarity = 0.713 M

    b) Determine the volume of this solution that would contain 3.93 moles of sodium chloride.

    Step 1: Data given

    Number of moles = 3.93 moles

    Molarity = 0.713 M

    Step 2: Calculate volume

    Volume = Moles / Molarity

    Volume = 3.93 mol/0.713 M

    volume = 5.51 L

    c) Determine the number of moles of sodium chloride in 22.45 L of this solution.

    Step 1: Data given

    Volume = 22.45 L

    Molarity = 0.713 M

    Step 2: Calculate number of moles

    Moles NaCl = Molarity * volume

    Moles NaCl = 0.713 * 22.45 L

    Moles NaCl = 16.01 moles
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