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 is added. How many grams of chloride ion were present in the 0.1000-g sample

0.016 grams of chloride ion were present in the 0.1000 grams of sample.

Explanation:

According to question, 9.00 mL of titrant was added to solution with 0.1000 grams of complex to reach the end point.

Molarity of the silver nitrate solution = 0.0500 M

Volume of the silver nitrate solution = V = 9.00 mL = 0.009 L

1 mL = 1000 L

Moles of silver nitrate = n

n = 0.00045 mol

According to 1 mole of silver nitrate reacts with 1 mol of chloride ion, then 0.00045 moles of silver nitrate will :

of chloride ions

Mass of chloride ions :

0.00045 mol × 35.5 g/mol = 0.016 g

0.016 grams of chloride ion were present in the 0.1000 grams of sample.

a) the particles of the chair are strongly held together.

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa