Moles Question?

Sodium carbonate forms a number of hydrates of the formula Na2CO3.XH20. A 3.01g sample of one of the hydrates was dissolved in water to make the solution 250cm3. In a titration, a 25cm3 portion of the solution required 24.4cm3 of 0.200 mol dm-3 HCL to complete the reaction.





Deduce the number of moles of Na2CO3 in 25cm3 of the Na2CO2 solution and hence deduce the number of moles of number of moles of Na2CO3 in the original 250cm3 solution.

Moles Question?
3.01g dissolved in 250 cc means 4*3.01=12.04g/1000cc solution;


In the titration :





N1V1 = N2V2


Here N1 = 0.2M=0.2N HCl;


V1 = 24.4 cc


N2 = ?


V2 = 25 cc


Hence,





N2 = 0.2*24.4/25





or N2 = 0.1952 N sodium carbonate;





Equivalent weight of sodium carbonate is =46+12+48 =106 /2 =53g;


the strength of sodium carbonate =0.1952*53 =10.3456g/dm-3;


Number of moles = 10.3456/106 =0.0976 M; or 0.0244M


Moles in 25cm3 of Na2CO3 =0.0976/40 =0.00244M


Actual Weight taken =12.04g/dm-3