Respuesta :
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.
The molarity of calcium chloride solution if 33.8 mL of it are needed to react with 41.2 mL of 0.877 M AgNO3 is 1.07M
HOW TO CALCULATE MOLARITY:
The molarity of calcium chloride solution can be calculated by using the following formula:
C1V1 = C2V2
Where;
C1 = concentration of calcium chloride
V1 = volume of calcium chloride
C2 = concentration of silver nitrate
V2 = volume of silver nitrate
According to this question, 33.8 mL of CaCl2 are needed to react with 41.2 mL of 0.877 M AgNO3.
33.8 × C1 = 0.877 × 41.2
33.8C1 = 36.13
C1 = 36.13 ÷ 33.8
C1 = 1.07M
Therefore, the molarity of calcium chloride solution if 33.8 mL of it are needed to react with 41.2 mL of 0.877 M AgNO3 is 1.07M.
Learn more at: https://brainly.com/question/2817451?referrer=searchResults
