Question

200 ml of 0.1 M AgNo3 is mixed with 300 ml of 0.2 M NaCl . calculate concentration of Cl- in resulting solution​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

0.08 m mol

Explanation:

you first find the no of moles of AgNO3 (20 m mol) and NaCl (60 m mol)

we see that AgNO3 is the limiting reagent

AgNO3 + NaCl --> AgCl + NaNO3

now, 20 mol of AgCl is formed

20 mol AgCl --> 20 mol NaCl

no of mol of NaCl left = 60-20 = 40

conc of Cl- = no of moles/total volume = 40/500 = 0.08

(here 500 ml is total volume, ie, 200+300)