Question

Ag+ + Cl- ↔ AgCl(s) 1.004 g of an impure mixture containing Cl- was dissolved to...

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?

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Homework Answers

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%

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