Sodium Carbonate crystals (27.82g) were dissolved in water and made up to 1dm^3. 25ml of the solution were neutralized by 48.8ml of HCl 0f 0.1M. Find n in the formulae of Na2CO3 crystals.
`Na_2CO_3+2HCl rarr 2NaCl+CO_2+H_2O`
Amount of HCl consumed `= 0.1/1000xx48.8`
Amount of `Na_2CO_3` reacted `= 1/2xx0.1/1000xx48.8 = 0.00244`
So we have 0.0244mol in 25ml of `Na_2CO_3` solution.
Amount of `Na_2CO_3` in `1dm^3 = 0.00244xx40 = 0.0976`
Mass of 0.0976 `Na_CO_3` moles `= 106xx0.0976 = 10.3456g`
Mass of water in `Na_2CO_3` crystals `= 27.82-10.3456 = 17.4744g`
Amount of water moles `= 17.474/18 ~~ 1`
So the answer is n = 1