Ask Question
8 December, 11:29

Calculate the volume in milliliters of a 1.1 M aluminum chloride solution that contains 125.0 g of aluminum chloride. Round your answer to significant digits.

+5
Answers (2)
  1. 8 December, 11:41
    0
    852 mL

    Explanation:

    Data of molarity means, moles of solute in 1L of solution

    Let's convert the mass of solute, into moles

    Mass / Molar mass = moles

    125 g / 133.33 g/mol = 0.937 moles

    In 1.1 moles of AlCl₃ are contained in 1L of solution

    0.937 moles of AlCl₃ are contained in (0.937 mol. 1L) / 1.1 mol = 0.852L

    Let's convert the L to mL

    0.852L. 1000 mL / 1L = 852 mL
  2. 8 December, 11:57
    0
    The volume of solution is 8.5*10^2 mL

    Explanation:

    Step 1: Data given

    Molarity of an aluminium chloride solution = 1.1 M

    Mass of aluminium chloride = 125.0 grams

    Molar mass of aluminium chloride = 133.34 g/mol

    Step 2: Calculate moles AlCl3

    Moles AlCl3 = mass AlCl3 / molar mass AlCl3

    Moles AlCl3 = 125.0 grams / 133.34 g/mol

    Moles AlCl3 = 0.9375 moles

    Step 3: Calculate volume of the solution

    Molarity = moles / volume

    Volume = moles / molarity

    Volume = 0.9375 moles / 1.1 M

    Volume = 0.852 L = 852 mL

    Volume = 8.5 * 10^2 mL

    The volume of solution is 8.5*10^2 mL
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Calculate the volume in milliliters of a 1.1 M aluminum chloride solution that contains 125.0 g of aluminum chloride. Round your answer to ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers