If you used Clearing in 2024, how did you find it?

Molarity, etc.

After a tumultuous year between high school and sixth form, I've started my A-levels four weeks late. If someone could go through this question with extensive methodology, it'd be beyond appreciated. I'm familiar with the general concepts but I haven't had time to refresh myself properly.

Sodium carbonate forms a number of hydrates of general formula Na2CO3.xH2O. A 3.01g sample of one of these hydrates was dissolved in water and the solution made up to 250cm³.In a titration, a 25.0cm³ portion of this solution required 24.3cm³ of 0.200 mol dm-³ hydrochloric acid for complete reaction.

The equation for this reaction is shown below.

Na2CO3 + 2HCl 2NaCl + 2H2O + CO2

Thus:

•

Calculate the number of moles of HCl in 24.3cm³ of 0.200 mol dm-³ hydrochloric acid.

•

Deduce the number of moles of Na2CO3 in 25.0cm³ of the Na2CO3 solution.

•

Hence deduce the number of moles of Na2CO3 in the original 250cm³ of solution.

•

Calculate the Mr of the hydrated sodium carbonate.

Also:

In an experiment, the Mr of a different hydrated sodium carbonate was found to be 250. Use this value to calculate the number of molecules of water of crystallisation, x, in this hydrated sodium carbonate, Na2CO3.xH2O.

Reply 1

Eau

Let's do the first part:

1. Using concentration = amount/volume (in units mol.dm-³ = mol/dm³),

The no. of moles of HCl used to neutralise the 25 cm³ of Na2CO3 = (24.3/1000)dm³ * 0.200 mol.dm-³ = 0.00486 mol

2. Looking at the reaction given, you can see that 1 mole of Na2CO3 reacts with 2 moles of HCl.

So, the number of moles of Na2CO3 in 25cm³ = 0.00486/2 = 0.00243 mol

3. No. of moles in the 250cm³ = 0.00243 * 10 = 0.0243 mol

4. This part may be a bit tricky:

First, you should know that mass = amount/Mr

Secondly, 1 mole of Na2CO3.xH2O generates 1 mole of Na2CO3.

so it follows that amount of Na2CO3.xH2O in 250 cm³ = 3.01/(23*2 + 12 + 16*3 + x(2+16)) = 3.01/(106 + 18x)

but this also equals 0.0243

Solving for x, you get x = 1.00 (2 d.p.)

Therefore, the hydrated form you used was Na2CO3.H2O, and the Mr of this is 124 mol/g.

Reply 2

Eau