Suppose you wanted to find out how many milliliters

of 1.0 M AgNO3 are needed to provide
169.88 g of pure AgNO3.
a. What is step 1 in solving the problem?
b. What is the molar mass of AgNO3?
c. How many milliliters of solution are needed?

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a. Step 1 in solving the problem is to calculate the number of moles of AgNO3 in 169.88 g of pure AgNO3.

b. To calculate the molar mass of AgNO3, we need to determine the mass of each element in one mole of AgNO3 and then sum them up. The molar mass is obtained by adding the atomic masses of all the atoms in the molecule. The atomic masses for each element can be found on the periodic table. Let's calculate the molar mass of AgNO3:

- Ag has an atomic mass of 107.87 g/mol
- N has an atomic mass of 14.01 g/mol
- O has an atomic mass of 16.00 g/mol (there are three oxygen atoms in AgNO3)

Now, let's calculate the molar mass of AgNO3:
Molar mass of AgNO3 = (107.87 g/mol) + (14.01 g/mol) + (16.00 g/mol * 3) = 169.87 g/mol

c. To determine the number of milliliters of 1.0 M AgNO3 solution needed, we need to convert the mass of AgNO3 to moles, and then use the molarity (mol/L) of the AgNO3 solution to find the volume in liters. We can use the formula:

volume (L) = moles / molarity

First, let's calculate the number of moles of AgNO3:
moles = mass / molar mass
moles = 169.88 g / 169.87 g/mol = 1.000 moles

Next, we can use the molarity of the AgNO3 solution to find the volume in liters:
volume (L) = moles / molarity
volume (L) = 1.000 moles / 1.0 mol/L = 1.000 L

To convert liters to milliliters, we multiply the volume in liters by 1000:
volume (mL) = 1.000 L * 1000 mL/L = 1000 mL

Therefore, 1000 milliliters of the 1.0 M AgNO3 solution are needed to provide 169.88 g of pure AgNO3.

a. Calculate mols AgNO3 needed. mols AgNO3 = grams/molar mass

b. 169.87
c. mols AgNO3 = M x L. You know M and mols solve for L and convert to mL.