KCl (1.4720 g) was made up to 250 cm3 solution. 25 cm3 reequired 21.13 cm3 AgNO3 for titration

KCl + AgNO3 ŕ AgCl + KNO3


what is the molarity of AgNO3

The question has two parts:-
Forst you need to work out the molarity of KCl
Use this equation

Amount (moles) = mass (grams) x volume (litres)

Remember to change the units of the volume form cm^3 to l


Second part knowing how many moles of KCL you can work out the moles of AgNO3
Take note that you only need a tenth of the initial soultion molarity you worked out
The equation shows the ratio of KCL:AGNO3 is 1:1

So if KCl works out ot be 10M (it's not) then you use 1M to react with 21.13 cm^3

25/21.13 = 1.18

1M /1.18 = 0.85M of AgNO3

Does that make sense?

If not I'll explain it a different way. In fact this website explains it differently but very nicely (they use less steps than me) Titration Calculations (http://www.webchem.net/notes/Periodicity/titration_calcs.htm)

thanks ever so much