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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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ilibee ilibee
wrote...
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11 years ago
How do I calculate the Molarity of each of the following solutions?
A) 4.3 mol of LiCl in 2.8 L solution
B) 22.6 g of C6H12O6 in 1.08 L solution
C) 45.5 mg of NaCl in 154.4 mL of solution
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2 Replies
Replies
wrote...
#1
11 years ago
M = moles /liter = mmoles/mL

A 4.3moles  /2.8 Liters =1.536 moles/Liter = 1.536 M

B a mole is molar mass in grams   so 22.6 grams of the compound C6H12O6 ( probably glucose )= 22.6 /180grams/per mole = 0.1255 moles
this mass placed in 1.08 Liters = 0.1255 moles/1.08 liters =
=0.1162moles/Liter = o.1162M

C  M= mmoles/mL  ( divide Moles andf liters by 1000)
    so 45.5 mg of NaCl/58.5 mg/mmol = 0.7777 mmoles
mmoles/mL = M soo 0.7777/154.4 mL = 0.0053 M
Answer accepted by topic starter
michelereemicheleree
wrote...
#2
Posts: 24
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11 years ago
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