We add excess Na2CrO4 solution to 68.0 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 0.270 grams. What is the molarity of the AgNO3 solution? Answer in units of M

Na2CrO4 + 2AgNO3 ==> Ag2CrO4

mols Ag2CrO4 = grams/molar mass = ?
mols AgNO3 = 1/2 that from the coefficients in the balanced equation.
M AgNO3 = mols AgNO3/L AgNO3.

.0769659509

To determine the molarity of the silver nitrate (AgNO3) solution, we need to use stoichiometry and the mass of the precipitate formed (Ag2CrO4).

First, let's write the balanced chemical equation for the reaction between silver nitrate (AgNO3) and sodium chromate (Na2CrO4):

2 AgNO3 + Na2CrO4 → Ag2CrO4 + 2 NaNO3

From the balanced equation, we can see that two moles of silver nitrate react with one mole of sodium chromate to form one mole of Ag2CrO4.

Now, let's calculate the moles of Ag2CrO4 formed:

moles of Ag2CrO4 = mass of Ag2CrO4 / molar mass of Ag2CrO4

The molar mass of Ag2CrO4 is determined by summing the atomic masses of all the elements in one formula unit:

Ag2CrO4 = (2 * atomic mass of Ag) + atomic mass of Cr + (4 * atomic mass of O)

Ag: atomic mass = 107.87 g/mol
Cr: atomic mass = 52 g/mol
O: atomic mass = 16 g/mol

Using these values, we can find:

molar mass of Ag2CrO4 = (2 * 107.87) + 52 + (4 * 16) = 331.87 g/mol

Now, substitute the given mass of Ag2CrO4 (0.270 grams) into the equation:

moles of Ag2CrO4 = 0.270 g / 331.87 g/mol

Calculating this value gives us the number of moles of Ag2CrO4.

Finally, since we know that two moles of AgNO3 react with one mole of Ag2CrO4, the number of moles of AgNO3 present in the solution is half the number of moles of Ag2CrO4.

moles of AgNO3 = (1/2) * moles of Ag2CrO4

To find the volume (in liters) of the AgNO3 solution, we convert the given volume of 68.0 mL to liters by dividing by 1000:

volume of AgNO3 solution = 68.0 mL / 1000 mL/L = 0.068 L

Finally, we can calculate the molarity (M) using the formula:

Molarity of AgNO3 = moles of AgNO3 / volume of AgNO3 solution

Substitute the calculated values to find the molarity of the AgNO3 solution.

To find the molarity of the AgNO3 solution, we need to use the balanced equation of the reaction between AgNO3 and Na2CrO4. This equation can be written as follows:

2 AgNO3 + Na2CrO4 -> Ag2CrO4 + 2 NaNO3

From this balanced equation, we can see that two moles of AgNO3 react with one mole of Na2CrO4 to form one mole of Ag2CrO4.

First, we need to calculate the number of moles of Ag2CrO4 formed. We have the mass of Ag2CrO4, which is given as 0.270 grams.

Next, we need to convert the mass of Ag2CrO4 to moles. We can do this using the molar mass of Ag2CrO4:

Ag (silver) has a molar mass of 107.87 g/mol.
Cr (chromium) has a molar mass of 52.00 g/mol.
O (oxygen) has a molar mass of 16.00 g/mol.

So, the molar mass of Ag2CrO4 is:
2(107.87 g/mol) + 52.00 g/mol + 4(16.00 g/mol) = 331.74 g/mol

Now, we can calculate the number of moles of Ag2CrO4:
0.270 g / 331.74 g/mol = 0.000814 mol

Since the balanced equation shows that two moles of AgNO3 react to form one mole of Ag2CrO4, the number of moles of AgNO3 is twice that of Ag2CrO4:

2(0.000814 mol) = 0.00163 mol

Finally, we can calculate the molarity of the AgNO3 solution.

Molarity (M) is defined as moles of solute divided by volume of solution in liters.

We know the volume of the AgNO3 solution is 68.0 mL, which is equal to 0.0680 L.

So, the molarity of the AgNO3 solution is:
0.00163 mol / 0.0680 L = 0.0239 M

Therefore, the molarity of the AgNO3 solution is 0.0239 M.