Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: blindsided on November 15, 2007, 08:56:56 PM
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What is the molarity of a solution that contains 12.5grams of NaHCO3 in 350mL solution?
so far I have:
12.5gNaHCO3 X 1mol/84gNaHCO3= .149molNaHCO3
I think next i should do:
.149mol/.35L= .43M NaHCO3
Not sure though.
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Dear Blindsided;
Yes! OK. - Please go ahead. (MolaRity = mol / Liter).
Good Luck!
ARGOS++
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ok. so: .43M NaHCO3/L is correct?
thanks bud
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Dear Blindsided;
Once again: Yes! - A little bit more exact: 0.425 mol NaHCO3 / Liter.
Good Luck!
ARGOS++