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mazda215 mazda215
wrote...
14 years ago
ok so i need help on these two questions..

1.Calculate the molarity of a solution that contains 5.35g NaCI in a 220 ML solution

2.What mass of Glucose C6H1206 is required to prepare 500 ML of a 2.50 M solution?


can anyone explain clearly on how to do these two problems? thanks if anyone answers
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wrote...
#1
Educator
14 years ago
For 1.

Find moles: NaCI = 5.35g / 58.443 g/mol (molar mass) = # moles
M = molality = # moles / 0.220 Kg of water = # (your answer)
mazda215 Author
wrote...
#2
14 years ago
so would it be .416M NaCl?....and any idea on the second question?
wrote...
#3
Educator
14 years ago
Exactly... Slight Smile Not sure about the second one though...
wrote...
#4
14 years ago
First find the molar mass of glucose... so find carbon * 6... hydrogen * 12 ... and oxygen * 6

so:

molar mass (g/mol) * 0.500 L * 2.50 mol / L

you get

molar mass (g/mol) * 0.500 L * 2.50 mol / L

Cool?
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