ohh man i wish i had a teacher like this..

originally...

4. a) 20.0 cm3 of a solution of ammonium chloride were boiled with 50 cm3 of 0.100 mol/dm3 sodium hydroxide solution. When no more ammonia gas was evolved, the resulting solution was allowed to cool and then titrated against a 0.100 mol/dm3 hydrochloric acid. If the volume of acid required was 28.2 cm3, what was the concentration of the ammonium chloride in g/dm3? Include two equations with state symbols in your calculation. Show all steps clearly. [9]

no. of mol of HCl = 0.100 * 0.0282 = 0.00282 mol
NaOH(aq) + HCl(aq)  NaCl(aq) + H20(l)
1 mol of NaOH reacts with 1 mol of HCl
no. of mol of NaOH titrated with HCl = 0.00282mol
no. of mol of NaOH initially = 0.100 * 0.050 = 0.0050mol
no. of mol of NaOH reacted with NH4Cl = 0.0050 – 0.00282 = 0.00218mol
NH4Cl(aq) + NaOH(aq)  NH3(g) + NaCl(aq) + H20 (l)
1 mol of NaOH reacts with 1 mol of NH4Cl
no. of mol of NH4Cl = 0.00218mol
mass of NH4Cl = 0.00218 * 18 = 0.03924
conc. Of ammonium chloride= 0.03924 / 0.020 = 1.962 g/dm^3

[[[ &&... ]]]

no. of mol of HCl = 0.100 * 0.0282 = 0.00282 mol yup
NaOH(aq) + HCl(aq)  NaCl(aq) + H20(l) yup
1 mol of NaOH reacts with 1 mol of HCl yup
no. of mol of NaOH titrated with HCl = 0.00282mol yup
no. of mol of NaOH initially = 0.100 * 0.050 = 0.00500mol yup
no. of mol of NaOH reacted with NH4Cl = 0.0050 – 0.00282 = 0.00218mol yup
NH4Cl(aq) + NaOH(aq)  NH3(g) + NaCl(aq) + H20 (l) yup
1 mol of NaOH reacts with 1 mol of NH4Cl yup
no. of mol of NH4Cl = 0.00218mol yup


mass of NH4Cl = 0.00218 * 18 = 0.03924 wrong! How you get this?
conc. Of ammonium chloride= 0.03924 / 0.020 = 1.962 g/dm^3 neh neh!


Actual workings (continued):

Volume of NH4Cl solution = 20.0cm3
Concentration of NH4Cl = 0.00218/0.020 = 0.109 mol/dm3
Concentration of NH4Cl = 0.109 * (14+4+35.5) = 5.83g/dm3

t a d a . = D !

(the side notes are his writings)
(i submitted my workings to my fren who send me the question via microsoft word)
zomg i want a teacher like this xD