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A sample contains 6.73 g of Na2CO3 in water to make a total of 250.0 mL of solution. What is the molarity of sodium carbonate in this solution and its mass-%? Use 1.00 g/mL as the density of the solution.

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  1. 12 May, 01:22
    0
    The molarity of Na2CO3 in the solution = 0.254 M

    The mass % of Na2CO3 is 2.69 %

    Explanation:

    Step 1: Data given

    Mass of Na2CO3 = 6.73 grams

    Volume of the solution = 250.0 mL

    Density = 1.00g/mL

    Molar mass of Na2CO3 = 105.99 g/mol

    Step 2: Calculate moles of Na2CO3

    Moles Na2CO3 = 6.73 grams / 105.99 g/mol

    Moles Na2CO3 = 0.0635 moles

    Step 3: Calculate molarity of Na2CO3

    Molarity = moles / volume

    Molarity = 0.0635 moles / 0.250 L

    Molarity = 0.254 M

    Step 4: Calculate mass of the solution

    Mass = volume * density

    Mass = 250 mL * 1.00g/mL

    Mass = 250 grams

    Step 5: Calculate mass %

    mass% = (6.73 g / 250 g) * 100%

    mass % = 2.69 %

    The mass % of Na2CO3 is 2.69 %
  2. 12 May, 01:38
    0
    Molarity of Na₂CO₃ = 0.25M

    % mass = 2.69

    Explanation:

    Molarity means mole of solute in 1L of solution

    Molar mass of solute (Na₂CO₃) = 105,98 g/m

    Moles = mass / molar mass → 6.73 g / 105.98 g/m = 0.0635 m

    Mol/L = [M]

    0.0635 mol/0.250L = 0.25M

    Density of solution = Solution mass / Solution volume

    1 g/ml = Solution mass / 250 mL → Solution mass is 250g

    % mass will be:

    In 250 g of solution we have 6.73 g of solute

    in 100 g of solution we have (100. 6.73) / 250 = 2.69
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