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:

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.

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.

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.

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.

to find the Mr, can’t you just doMr= mass/moles Mr= 3.01/0.0243= 123.8 gmol-1

Swear down x is 10 explain again pls

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