Asked by denise
How many moles of solute are in the following solutions?
Part A
250mL of 0.26M acetic acid, CH3CO2H
2.90L of 0.31M NaOH
980mL of 3.5M nitric acid, HNO3
Part A
250mL of 0.26M acetic acid, CH3CO2H
2.90L of 0.31M NaOH
980mL of 3.5M nitric acid, HNO3
Answers
Answered by
MathMate
molar mass of CH3COOH
=(12+3+12+16+16+1)
=60
mass of acetic acid in 250 mL of 0.26M
=(250/1000) L * 0.26*60
= 3.9 g
=(12+3+12+16+16+1)
=60
mass of acetic acid in 250 mL of 0.26M
=(250/1000) L * 0.26*60
= 3.9 g
Answered by
DrBob222
What Mathmate gave you is correct for grams acetic acid in the solution. The problems ask for mols and that is obtained by mols = M x L = ? (in this case just omit the 60).
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