Answer:
0.47 M
Explanation:
The concentration of the solution can be calculated using the following equation:
[tex] C = \frac{\eta}{V} = \frac{m}{M*V} [/tex]
Where:
V: is the volume of the solution = 68.6x10⁻² L
η: is the moles of cobalt (II) sulfate
m: is the mass of cobalt (II) sulfate = 89.94 g
M: is the molar mass of cobalt (II) sulfate = 281.103 g/mol
The concentration of cobalt (II) sulfate is:
[tex] C = \frac{m}{M*V} = \frac{89.94 g}{281.103 g/mol*68.6 \cdot 10^{-2} L} = 0.47 mol/L [/tex]
We used the molar mass of the cobalt (II) sulfate heptahydrate (281.103 g/mol) since it is one of the most common salts of cobalt.
Therefore, the concentration of a solution of cobalt (II) sulfate is 0.47 M (assuming that the cobalt (II) sulfate is heptahydrate).
I hope it helps you!
Answer:
[tex]M=0.467M[/tex]
Explanation:
Hello,
In this case, cobalt (II) molar mass is 280.996 g/mol since it is about heptahydrate, so we compute the moles:
[tex]n=89.94g*\frac{1mol}{280.996g}= 0.32mol[/tex]
Then, since we need the volume in litres only:
[tex]V=68.6cL*\frac{1L}{100cL} =0.686L[/tex]
Finally, we compute the molarity:
[tex]M=\frac{0.32mol}{0.686L}\\ \\M=0.467M[/tex]
So your answer should be refined.
Best regards.
