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3 years ago

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

1 answer:

3 0

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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3 years ago

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$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

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$3.89=-\log[H_3O^+]$

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Hence, the hydronium ion concentration in the solution is $1.29\times 10^{-4}M$

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