Question

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

Answer 1

Answer:

$M=0.467M$

Explanation:

Hello,

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

$n=89.94g*\frac{1mol}{280.996g}= 0.32mol$

Then, since we need the volume in litres only:

$V=68.6cL*\frac{1L}{100cL} =0.686L$

Finally, we compute the molarity:

$M=\frac{0.32mol}{0.686L}\\ \\M=0.467M$

So your answer should be refined.

Best regards.

Answer 2

Answer:

0.47 M

Explanation:

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

$C = \frac{\eta}{V} = \frac{m}{M*V}$

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:

$C = \frac{m}{M*V} = \frac{89.94 g}{281.103 g/mol*68.6 \cdot 10^{-2} L} = 0.47 mol/L$

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!