Question

Ag⁺ + Cl^- ↔ AgCl_(s)

1.004 g of an impure mixture containing Cl^- was dissolved to 100.0 mL. 10.00 mL of this solution required 22.97 mL of 0.05274 M AgNO₃ to reach end point. What is the weight percent of Cl^- in the unknown solid?

Answer 1

We first use the formula M1V1 = M2V2 to find the molarity of the impure solution titrated against AgNO3. M2 and V2 represent the molarity and volume of AgNO3 respectively.

M1*10 = 0.05274*22.97

M1 = 0.1211 M

The molarity of the impure solution is 0.1211 M

Molarity = number of moles of solute/ volume in L

0.1211= number of moles /0.1L

Number of moles = 0.01211

Number of moles = given mass/ molar mass

Molar mass = 1.004/0.01211 = 82.9 g

Molar Mass of Cl- = 35.453 g

Weight percentage of Cl- = (mass of cl/mass of the unknown compound)*100

=( 35.453/82.9 )*100 = 42.76%