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What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2
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Dec 6, 2017
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What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2
Chemistry
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Aryanm
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Molar mass CaCl₂ = 111.0 g/mol
number of moles:
n = mass of solute / molar mass
n = 85.3 / 111.0
n = 0.7684 moles of CaCl₂
M = n / V
0.788 M =
0.7684 / V
V = 0.7684 / 0.788
V = 0.97512 L
hope this helps!
Austen Chongpison
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Dec 11, 2017
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Austen Chongpison
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