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24 August, 10:55

A solution is created by dissolving 13.0 grams of ammonium chloride in enough water to make 325 mL of solution. How many moles of ammonium chloride are present in the resulting solution?

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Answers (2)
  1. 24 August, 13:13
    0
    We have 0.243 moles of ammonium chloride is the solution

    Explanation:

    Step 1: Data given

    Ammonium chloride = NH4Cl

    Mass of ammonium chloride = 13.0 grams

    Volume of water = 325 mL = 0.325 L

    Molar mass of ammonium chloride

    ⇒ Molar mass N = 14.0 g/mol

    ⇒ Molar mass of H = 1.01 g/mol : 4*1.01 = 4.04 g/mol

    ⇒Molar mass of Cl = 35.45 g/mol

    Molar mass NH4Cl = 53.49 g/mol

    Step 2: Calculate moles ammonium chloride

    Moles NH4Cl = mass NH4Cl / molar mass NH4Cl

    Moles NH4Cl = 13.0 grams / 53.49 g/mol

    Moles NH4Cl = 0.243 moles

    We have 0.243 moles of ammonium chloride is the solution
  2. 24 August, 13:55
    0
    0.243 moles of ammonium chloride are present in the resulting solution

    Explanation:

    We determine our compounds of the solution:

    Solute: NH₄Cl

    The solvent is water

    Solution's volume: 325 mL

    If we want to know, how many moles of ammonium chloride are present in the resulting solution we may just convert the solute's mass to moles.

    Molar mass NH₄Cl = 53.45 g/mol

    Moles of NH₄Cl = Mass of NH₄Cl / Molar mass NH₄Cl

    13 g / 53.45 g/mol = 0.243 moles
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