- #1

Physiona

- 131

- 9

## Homework Statement

Sodium carbonate exists in hydrated form, Na

_{2}CO3.xH2O, in the solid state. 3.5 g of a sodium carbonate sample was dissolved in water and the volume made up to 250

^{cm3}. 25.0

^{cm3}of this solution was titrated against 0.1

^{ moldm-3}HCl and 24.5 cm3 of the acid were required. Calculate the value of x given the equation:

Na

_{2}CO

_{3 }+ 2HCl --> 2NaCl + CO2 + H2O

## Homework Equations

NO. of Moles = Conc. x volume/1000

## The Attempt at a Solution

I have attempted this question in this way,

Moles of HCI = 0.1 x 24.5 / 1000 = 0.00245

Then looking at the molar ratio, I see 1:2 ratio of Sodium carbonate and HCI.

So, Moles of Na2CO3, is 0.00245/2 = 0.001225 moles in 25

^{cm3}

Then I did the NO. of moles for Na

_{2}CO

_{3}in 250

^{cm3}so 250

^{cm3}/25

^{cm3}= 10 which is the multiplier, then i did 10 x 0.001225 = 0.01225 mol

I did the Molar Mass of Na

_{2}CO3 which is 3.5/0.01225 = 285.7142.. which rounds to 286.

I assumed the Mr of Na2CO3.xH2O is 286, as the mark scheme assumed that

I don't know what to do after this, can someone please help me, as this is Homework!

Thank you!