Asked by Charlie
How would one work through this problem? -Calculate the number of liters of 1.25M solution that can be made from 6.400g of barium nitrate. -
I know that it's just stoichiometry, but I'm confused. The formula for molarity is the number of moles of the solute divided by liters of solution, so wouldnt the set up be 1.25=6.4/x ? In which case x would be .195 but I doubt it's correct
I know that it's just stoichiometry, but I'm confused. The formula for molarity is the number of moles of the solute divided by liters of solution, so wouldnt the set up be 1.25=6.4/x ? In which case x would be .195 but I doubt it's correct
Answers
Answered by
DrBob222
How many mols Ba(NO3)2 do you have?
That's mols = grams/molar mass = ?
approx 0.02 but that's just an estimate.
Then M = mols/L
You know M(1.25) and mols (estimated 0.02), solve for L.
That's mols = grams/molar mass = ?
approx 0.02 but that's just an estimate.
Then M = mols/L
You know M(1.25) and mols (estimated 0.02), solve for L.
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