You can go about this two ways.
(g solute/ mL)*100 = %
(g CaCO3/250 mL)*100 = 5%
Solve for g CaCO3.
The other way is that 5% w/v means 5g CaCO3 in 100 mL, you want 250 so
5g CaCO3 x (250/100) = g needed for 250 mL instead of 100 mL.
By the way I don't think this is possible because CaCO3 is not all that soluble.
To make 250 ml of a 5%(w/v) calcium carbonate solution, how many grams of calcium carbonate will I need?
2 answers
I believe it's hypothetical. Just some simple review for my finals and this one slipped my mind. Thank you so much. 12.5g is needed for 250mL