Calculate the molarity of a calcium chloride solution if 33.8 mL of it are needed to react with 41.2 mL of 0.877 M AgNO3. Report your answer in standard notation with three significant figures, and do not include units in your answer.

0.465 M IS THE MOLARITY OF CALCIUM CHLORIDE SOLUTION.

Explanation:

Volume of CaCl2 = 33.8 mL = 33.8 / 1000 L = 0.0338 L

Volume of AgNO3 = 41.2 mL = 0.0412 L

Molarity of AgNO3 = 0.877 M

First is to write and balance out the equation for the reaction

2AgNO3 + CaCl2 > 2AgCl + Ca(NO3)2

From the reaction, we could see that 2 moles of AgNO3 combines with 1 mole of CaCl2 to produce 12 moles of AgCl and 1 mole of Ca(NO3)2.

In volmetric analysis, the relationship between molarity and volume of acids and bases is shown in the equation below:

MaVa / Mb Vb = Na / Nb

Ma = molarity of acid = 0.877 M

Va = volume of acid = 0.0412 L

Mb = molarity of base = unknown

Vb = volume of base = 0.0338 L

Na = number of moles of acid = 2

Nb = number of mole of base = 1

Rearranging the formulae, making Mb the subject of the equation, we get:

Mb = Ma Va Nb / Vb Na

Mb = 0.877 * 0.0412 * 1 / 0.0338 * 2

Mb = 0.0361 / 0.0776

Mb = 0.465 M

So therefore, the molarity of calcium chloride solution if the variables given were used is 0.465 M.

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