We add excess Na2CrO4 solution to 65.0 ml of a solution of silver nitrate to form insoluble solid Ag2CrO4. When it has been dried and weighed, the mass of Ag2CrO4 is found to be 0.640 grams. What is the molarity of the AgNO3 solution? Answer in units of M
To determine the molarity of the AgNO3 solution, we need to first calculate the moles of Ag2CrO4 formed.
First, calculate the molar mass of Ag2CrO4:
Ag: 107.87 g/mol
CrO4: 116.96 g/mol
Molar mass of Ag2CrO4 = 2(107.87) + 116.96 = 332.7 g/mol
Now calculate the moles of Ag2CrO4 formed:
moles = mass / molar mass = 0.640 g / 332.7 g/mol = 0.00192 mol
Since Ag2CrO4 was formed by the reaction between AgNO3 and Na2CrO4, the moles of AgNO3 is equal to the moles of Ag2CrO4 produced.
Now, calculate the volume of the AgNO3 solution:
65.0 ml = 0.065 L
Finally, calculate the molarity of the AgNO3 solution:
Molarity = moles / volume = 0.00192 mol / 0.065 L = 0.0295 M
Therefore, the molarity of the AgNO3 solution is 0.0295 M.