Question

30.00cm3 of a solution (B1) containing 60 g/dm^3 NaCl/Na2CO3 mixture is placed in a 250cm^3 graduated flask using a burette. The contents of the flask are made up to the 250 cm^3 mark using distilled water (this solution is now call B2). How many moles of Na2CO3 are there in 25cm^3 of B2.

*no additional information was given

Answer 1

Ans. Since no information about relative amount of NaCl and Na₂CO₃ is provides, it’s assumed that the mixture is a 1:1 ratio by mass.

# Given, the solution is 60 g / dm³ of NaCl/ Na₂CO₃

Mass of Na₂CO₃ in solution B = (½) x Total mass of mixture

                                                = (½) x 60.0 g

                                                = 30.0 g

Moles of Na₂CO₃ = Mass / Molar mass

                                    = 30.0 g / (105.988736 g/ mol)

                                    = 0.28304895 mol

Molarity of Na₂CO₃ in solution B = Moles of Na₂CO₃ / Volume of solution in liters

                                    = 0.28304895 mol / 1.0 L                            ; [1 dm³ = 1.000 L]

                                    = 0.28304895 M

# Preparation of solution B2: 30.0 mL of solution B1 is made upto 250.0 mL with distilled water.

Now, using C1V1 (B1 solution) = C2V2 (B2 solution)

            Or, 0.28304895 M x 30.0 mL = C2 x 250.0 mL

            Or, C2 = (0.28304895 M x 30.0 mL) / 250.0 mL

            Hence, C2 = 0.033965874 M

Therefore, [Na₂CO₃] in solution B2 = 0.033965874 M

# Desired volume of solution B2 = 25.0 cm³ = 25.0 mL = 0.025 L

Now,

Moles of Na₂CO₃ in 25.0 mL solution B2 = [Na₂CO₃] in soln. B2 x Desired volume in Liters

                                    = 0.033965874 M x 0.025 L

                                    = 0.00084914685 mol

Hence, moles of Na₂CO₃ in 25.0 mL solution B2 = 8.4915 x 10^-4 moles