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Stella [2.4K]
3 years ago
11

What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2?

Chemistry
1 answer:
Anna11 [10]3 years ago
5 0
Molar mass  CaCl₂ =  110.98 g/mol

Number of moles:

1 mole CaCl₂ ---------> 110.98 g
n mole CaCl2 ---------> 85.3 g

n = 85.3 / 110.98

n = 0.7686 moles of CaCl₂

Volume = ?

M = n / V

0.788 =  0.7686 / V

V = 0.7686 / 0.788

V = 0.975 L

hope this helps!
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