Chemistry : determine masses of anhydrous sodium carbonate

Question

Given the Relative Atomic Masses, C = 12, O = 16, Na = 23, determine the mass of anhydrous sodium carbonate needed to make up the following solutions.

250 cm3 of 0.2 mol dm-3 solution

500 cm3 of 0.05 mol dm-3 solution

100 cm3 of 0.5 mol dm-3 solution

Answer 1

Amount (mol) = concentration (mol dm-3) x volume (dm3)

1 dm3 is sometimes called 1 L

First example: 250 cm3 x 1 dm3/1,000 cm3 x 0.2 mol dm-3 = 0.05 mol

mol x relative molar mass (g/mol) = mass

Molar mass of anhydrous Na2CO3: 2 x 23 + 12 + 3 x 16.

Make sure you understand why what I am saying works.  It is really just bookkeeping of amounts, but you need to get it out of the way before getting on to the interesting stuff.

Answer 2

Mr(Na2CO3) =

Na x 2 = 23 x 2 = 46

C x 1 = 12 x 1 = 12

0 x3 = 16 x 3 = 48

46 + 12 + 48 = 106

As 250 cm^3  is 1/4 (0.25) of 1 dm^3

Then 0.25 x 0.2 = 0.05 mol is required.

Hence mass = 0.05 x 106 = 5.3 g

Similarly

As 500 cm^3  is 1/2 (0.5) of 1 dm^3

Then 0.5 x 0.05 = 0.025 mol is required.

Hence mass = 0.025 x 106 = 2.65 g

Similarly

As 100 cm^3  is 1/10 (0.1) of 1 dm^3

Then 0.1 x 0.5 = 0.05 mol is required.

Hence mass = 0.05 x 106 = 5.3 g

Hope that helps !!!!!