it is desired to prepare 250.0mL of a standard solution having a concentration of 0.1000M AgNo3. How many grams of a sample of AgNO3 of 98.8% purity are required for this purpose?
M = moles/L
0.1 = moles/0.250 L
solve for moles.
moles = grams/molar mass
You know moles and molar mass, solve for grams (100% pure) AgNO3.
grams impure AgNO3 x 0.988 = grams pure AgNO3 you want or rearrange to
grams pure AgNO3/0.988 = grams you must weigh out to get grams AgNO3 you need.
the answer in the book was 4.3 grams
but if you do that you get
169.8731/0.988 which is 171.94 g
i don't get it
To find out how many grams of AgNO3 are required, we need to first calculate the number of moles of AgNO3 needed for the desired concentration, and then use the molar mass and purity of AgNO3 to calculate the mass.
Step 1: Calculate the number of moles of AgNO3 needed.
We can use the formula:
moles = concentration (M) x volume (L)
Given:
concentration (M) = 0.1000 M
volume (L) = 250.0 mL
First, we need to convert the volume to liters:
volume (L) = 250.0 mL ÷ 1000 = 0.250 L
Now we can calculate the number of moles:
moles = 0.1000 M x 0.250 L = 0.025 mol
Step 2: Calculate the mass of AgNO3 needed.
To calculate the mass, we can use the formula:
mass (g) = moles x molar mass (g/mol)
Given:
purity of AgNO3 = 98.8%
First, we need to find the molar mass of AgNO3:
Ag (atom) = 107.87 g/mol
N (atom) = 14.01 g/mol
O (atom) = 16.00 g/mol
molar mass (AgNO3) = (107.87 + 14.01 + (3 x 16.00)) g/mol = 169.87 g/mol
Now we can calculate the mass of AgNO3 needed:
mass (g) = 0.025 mol x 169.87 g/mol = 4.2475 g
Finally, considering the purity of the sample AgNO3, we can calculate the actual mass required:
actual mass (g) = mass (g) ÷ purity
actual mass (g) = 4.2475 g ÷ 0.988 = 4.301 g
Therefore, approximately 4.301 grams of AgNO3 of 98.8% purity are required to prepare the desired solution.