Question

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

Answer 1

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!