What Is the Molar Ratio of Water in Hydrated Sodium Carbonate?

[nobahar]

Homework Statement

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 was made up to 250cm³. In a titration, 25cm³ portion of this solution required 24.3cm³ of 0.2 mol dm^-3 hydrochloric acid for complete reaction.

Homework Equations

Na₂CO₃ + 2HCl is the important bit

The Attempt at a Solution

I got the number of moles of HCl to be 4.86 x 10^-3 mol; therefore the number of moles of Na₂CO₃ to be 2.43 x 10^-3 mol in 25cm³ and so 0.0243 mol in the orginal 250cm³.
This must be the number of moles of Na₂CO₃ in the original 3.01g. 0.0243 moles of Na₂CO₃ weighs 2.5758g, and so the water must make up 0.4342g (derived from 3.01g - 2.5758g). The number of moles of H₂O is therefore 0.024. I figured that the number of moles of Na₂CO₃ to H₂O is approximately 1, and so the orginal formula must be Na₂CO₃.H₂O (i.e. x is 1).
The M_r is therefore106+(1 x 18) = 124g/mol.
Is this correct?
Thanks in advance.

 

[nobahar]

I fear may question may appear too long. This is the important part:

This must be the number of moles of Na₂CO₃ in the original 3.01g. 0.0243 moles of Na₂CO₃ weighs 2.5758g, and so the water must make up 0.4342g (derived from 3.01g - 2.5758g). The number of moles of H₂O is therefore 0.024. I figured that the number of moles of Na₂CO₃ to H₂O is approximately 1, and so the orginal formula must be Na₂CO₃.H₂O (i.e. x is 1).
The M_r is therefore106+(1 x 18) = 124g/mol.

Thanks in advance.

 

[Borek]

Looks OK to me.

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[nobahar]

Thanks Borek.