We add excess Na2CrO4 solution to 42.0 mL
of a solution of silver nitrate (AgNO3) to form
insoluble solid Ag2CrO4. When it has been
dried and weighed, the mass of Ag2CrO4 is
found to be 0.590 grams. What is the molarity
of the AgNO3 solution?
Answer in units of M
I worked one like this (only the numbers are different) earlier today.
http://www.jiskha.com/display.cgi?id=1382130924
To find the molarity of the AgNO3 solution, we need to determine the number of moles of Ag2CrO4 formed and then use stoichiometry to relate it to the moles of AgNO3.
First, let's calculate the number of moles of Ag2CrO4 formed. We can use the molar mass of Ag2CrO4 to convert its mass to moles.
Given:
Mass of Ag2CrO4 = 0.590 grams
The molar mass of Ag2CrO4 is calculated as follows:
Molar mass of Ag = 107.87 g/mol
Molar mass of Cr = 52.00 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of Ag2CrO4 = (2 * Molar mass of Ag) + Molar mass of Cr + (4 * Molar mass of O)
Molar mass of Ag2CrO4 = (2 * 107.87 g/mol) + 52.00 g/mol + (4 * 16.00 g/mol)
Molar mass of Ag2CrO4 = 331.87 g/mol
Now we can calculate the number of moles using the formula:
moles = mass / molar mass
moles of Ag2CrO4 = 0.590 g / 331.87 g/mol
Next, we need to determine the stoichiometric ratio between Ag2CrO4 and AgNO3. From the chemical equation, we can see that 1 mole of Ag2CrO4 is formed for every 2 moles of AgNO3.
Therefore, the moles of AgNO3 can be calculated as follows:
moles of AgNO3 = (moles of Ag2CrO4) * (2 moles of AgNO3 / 1 mole of Ag2CrO4)
Now, we can calculate the volume in liters using the formula:
Molarity = moles / volume
Volume of AgNO3 solution = Volume of Ag2CrO4 solution = 42.0 mL = 0.0420 L
Finally, we can calculate the molarity of the AgNO3 solution:
Molarity = (moles of AgNO3) / (volume of AgNO3 solution)
Note: Make sure to convert the volume from mL to L before calculating the molarity.
Please substitute the values into the above equations to obtain the final answer.
To find the molarity of the AgNO3 solution, we need to use the balanced chemical equation for the reaction between Na2CrO4 and AgNO3 to form Ag2CrO4:
2 AgNO3 + Na2CrO4 → Ag2CrO4 + 2 NaNO3
From the balanced equation, we can see that the ratio of AgNO3 to Ag2CrO4 is 2:1. This means that for every 2 moles of AgNO3, we will get 1 mole of Ag2CrO4.
First, let's calculate the number of moles of Ag2CrO4 formed:
Given mass of Ag2CrO4 = 0.590 grams
To convert grams to moles, we need the molar mass of Ag2CrO4. The molar mass of Ag is approximately 107.87 g/mol, Cr is approximately 52.00 g/mol, and O is approximately 16.00 g/mol.
Molar mass of Ag2CrO4 = (2 * 107.87 g/mol) + 52.00 g/mol + (4 * 16.00 g/mol)
= 331.74 g/mol
Number of moles of Ag2CrO4 = mass / molar mass
= 0.590 g / 331.74 g/mol
Now, let's calculate the number of moles of AgNO3 used:
Since the ratio of AgNO3 to Ag2CrO4 is 2:1, the number of moles of AgNO3 used is half the number of moles of Ag2CrO4.
Number of moles of AgNO3 = (0.590 g / 331.74 g/mol) / 2
Next, we'll calculate the volume of the AgNO3 solution:
Given volume of AgNO3 solution = 42.0 mL
We need to convert the volume to liters by dividing by 1000:
Volume of AgNO3 solution = 42.0 mL / 1000 mL/L
Now, we have all the required information to calculate the molarity of the AgNO3 solution:
Molarity (M) = number of moles / volume (in liters)
= (0.590 g / 331.74 g/mol) / 2 / (42.0 mL / 1000 mL/L)
Simplifying the equation, we get:
Molarity (M) = (0.590 / 331.74) / 2 / (42.0 / 1000)
= (0.001776) / (0.042)
Now, we can calculate the molarity:
Molarity (M) ≈ 0.0421 M
Therefore, the molarity of the AgNO3 solution is approximately 0.0421 M.