Question

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.

Answer 1

The molarity of Na2CO3 in the solution = 0.254 M

The mass % of Na2CO3 is 2.69 %

Step-by-step 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 %

Answer 2

Molarity of Na₂CO₃ = 0.25M

% mass = 2.69

Step-by-step 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