hello everyone :D
thanks to xinle, i have realised my mistake. i have re-attempted the question in such a way that it is similar to one of the solutions in our chemistry worksheet. please help me point at my mistake again! :D
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20.0 cm3 of a solution containing NaCl and KCl gave, on evaporation to dryness, 0.180 g of the mixed chlorides.
20.0 cm3 of the same solution gave 0.370 g of AgCl on treatment with a slight excess of AgNO3 solution.
Calculate, for the original solution, the mass per dm3 of (a) NaCl and (b) KCl

Let the amount of NaCl be yg.
Let the amount of KCl be (0.180-y)g.

AgNO3 + NaCl ---> AgCl +NaNO3
AgNO3 + KCl ---> AgCl +KNO3

Number of moles of AgCl = 0.370/ (108+35.5)
= 2.5784*10^-3 (5s.f.) STORE THIS VALUE IN THE CALCULATOR!

Molar mass of NaCl = 58.5 g/mol
Molar mass of KCl = 74.6 g/mol

AgCl is produced by the reaction of (y/58.5) moles of NaCl and [(0.180-y)/74.6] moles of KCl.

(y/58.5) + [(0.180-y)/74.6] = 2.5784*10^-3
74.6y + 10.53 - 58.5y = 11.252 (5s.f.) STORE THIS VALUE INTO CAL.
16.1y = 0.72238 (5s.f.) STORE THIS VALUE INTO CAL.
y = 0.044869 (5s.f.) STORE THIS VALUE INTO CAL.

Therefore,
Mass of NaCl = 0.044869 g.
Mass of KCl = 0.180-0.044869
= 0.13513 g. STORE THIS VALUE INTO CAL.


Concentration of NaCl = 0.044869 / (20/1000)
= 2.24 g/dm3 (3s.f.)
Concentration of KCl = 0.13513 / (20/1000)
= 6.76 g/dm3 (3s.f.)

Let's hope this is right this time.

love, shi tian