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mazda215 mazda215
wrote...
13 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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Replies
wrote...
Educator
13 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...
13 years ago
so would it be .416M NaCl?....and any idea on the second question?
wrote...
Educator
13 years ago
Exactly... Slight Smile Not sure about the second one though...
wrote...
13 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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