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
2 answers
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.
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.
1 answer