johnarm23 johnarm23

27-04-2016
Chemistry

Answered

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

Verified Answer :

Willchemistry Willchemistry

27-04-2016

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!

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