Question

What is the concentration of a solution formed by mixing 89.94 grams of cobalt (II)

sulfate in enough water to make 68.6 cL solution?


I keep getting .84 M. Can anyone explain how to get the right answer?

Answer 1

`M=0.467M`

Step-by-step explanation:

Hello,

In this case, cobalt (II) molar mass is 280.996 g/mol since it is about heptahydrate, so we compute the moles:

`n=89.94g*(1mol)/(280.996g)= 0.32mol`

Then, since we need the volume in litres only:

`V=68.6cL*(1L)/(100cL) =0.686L`

Finally, we compute the molarity:

`M=(0.32mol)/(0.686L)\\ \\M=0.467M`

So your answer should be refined.

Best regards.

Answer 2

0.47 M

Step-by-step explanation:

The concentration of the solution can be calculated using the following equation:

`C = (\eta)/(V) = (m)/(M*V)`

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:

`C = (m)/(M*V) = (89.94 g)/(281.103 g/mol*68.6 \cdot 10^(-2) L) = 0.47 mol/L`

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!