16 g KNO3/molar mass KNO3 = mols KNO3.
M of the diluted solution is
mL1 x M1 = mL2 x M2
44.8 x 7.0 = 1100 x M2
Solve for M2.
Then M = mols/L or L = mols/M. You know mols in the 16 g, you know M of the diluted solution, solve fo4 L of the diluted solution.
A 44.8−mL sample of a 7.0M KNO3 solution is diluted to 1.10L What volume of the diluted solution contains 16.0g of KNO3?
Would I do the same thing for this one or would does it have an extra step it it.
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