If 10g of AgNO3 is available, what volume of 0.25 M AgNO3 solution can be prepared?
0.25 M = 0.25 M = 0.25 L
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1 L
Do I have to do anything with the 10g or is it just extra info
0.25 M/1L = 0.25 L
Of course the 10 g is real information. It's the same thing as how many ice cream cones can you buy for $10.00 if each cone costs $1.00?
You want 0.25 M AgNO3 which is 0.25 mols AgNO3/L solution.
So how many mols AgNO3 do you have in 10 g AgNO3. That's mols = grams/molar mass = 10.00/approx 170 (you need a better number than that) = approx 0.06 mols. Then 0.25 M = mols/L
0.25 = 0.05/L solution. Solve for L solution. That will be approx 0.2 L total solution. All of those numbers I've used are estimates but they are close estimates.
The 10g of AgNO3 is not extra information, it is essential to calculate the volume of the 0.25 M AgNO3 solution that can be prepared. To do this, you need to use the formula:
moles = mass / molar mass
First, determine the molar mass of AgNO3 by adding up the atomic masses of silver (Ag), nitrogen (N), and three oxygen (O) atoms. The atomic masses of Ag, N, and O are approximately 107.87 g/mol, 14.01 g/mol, and 16.00 g/mol respectively. So the molar mass of AgNO3 is approximately:
(107.87 g/mol) + (14.01 g/mol) + (3 * 16.00 g/mol) = 169.87 g/mol
Next, convert the given mass of AgNO3 (10g) into moles by dividing it by the molar mass:
moles = 10g / 169.87 g/mol
Now that you have the number of moles, you can calculate the volume of the solution using the definition of molarity:
molarity (M) = moles / volume (L)
Rearranging the formula, we get:
volume (L) = moles / molarity (M)
Substituting the values we have:
volume (L) = (10g / 169.87 g/mol) / 0.25 M
Simplifying the equation, we find:
volume (L) = 0.059 L
Therefore, the volume of 0.25 M AgNO3 solution that can be prepared from 10g of AgNO3 is approximately 0.059 liters, or 59 milliliters.