Question

# Addition of AgNO3 to aqueous solutions of the complex results in a cloudy white precipitate, presumably...

Addition of AgNO3 to aqueous solutions of the complex results in a cloudy white precipitate, presumably AgCl. You dissolve 0.1000 g of the complex in H2O and perform a precipitation titration with 0.0500 M AgNO3 as the titrant. Using an electrode that is sensitive to [Ag+], you reach the endpoint after 9.00 mL of titrant are added. How many grams of chloride ion were present in the 0.1000-g sample?

(The answer is E, Could anyone explain it ?

 A) 4.50 ´ 10–4 B) 5.00 ´ 10–3 C) 1.77 ´ 10–3 D) 6.38 ´ 10–2 E) 1.60 ´ 10–2

Reaction occurs:

Ag+ (aq) + Cl- (aq) = AgCl (s)

According to above equation:

n(Ag+ ) = n(Cl-)

n-moles

From titration:

n=c*V

c-concentration, mol/L

V-volume, L

n(Ag+ ) = n(AgNO3) = 0.05 mol/L * 0.009 L (Volume added to endpoint in Liters) = 0.00045 mol = 4.5*10-4 moles

As mentioned above:

n(Cl-) = n(Ag+ ) = n(AgNO3) = 4.5*10-4 moles

m = n*M

m-mass, in g

M-molar mass in g/mol, for anion Cl- 35.45 g/mol. So

m(Cl-) = 4.5*10-4 moles * 35.45 g/mol = 1.595*10-2 g = 1.6*10-2 g

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