# Important Stoichiometry question, just need formula or help and answer checking [closed]

How many grams of Cobalt(II) chloride $$\ce{CoCl2}$$ is needed to make $$\pu{0.645 L}$$ with a concentration of $$\pu{1.25 mol\:L^{-1}}$$? (the molar mass of anhydrous $$\ce{CoCl2}$$ is $$\pu{129.839 g\: mol^{-1}}$$, I believe)

## closed as off-topic by andselisk♦, A.K., Todd Minehardt, user55119, airhuffMar 19 at 6:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

• This is definitely a homework question. You must need to show your effort to solve it. – Mathew Mahindaratne Mar 19 at 2:03
• well, could you show me the formula possibly? I don't know how to get help with this question and I don't know how to do it... – Nam Huy Mar 19 at 2:05
• First, you need to find the amount of $\ce{CoCl2}$ in $\pu{mols}$ to make $\pu{0.645 L}$ of $\pu{1.25 M}$ solution. – Mathew Mahindaratne Mar 19 at 2:12
• I believe it is .807, what do I do after this? – Nam Huy Mar 19 at 2:18
• Correct answer is $\pu{0.806 mol}$. So, do you remember, $g= \pu{mol} \times \text{molar mass}$? – Mathew Mahindaratne Mar 19 at 2:24

Remember the formula $$c = \frac{n}{v}$$ where $$c$$ is the molarity, which is moles of solute per liter of solution. You can rearrange it to get $$c \times v = n$$. Thus in your case $$\pu{1.25M} \times \pu{0.645L} = \pu{0.806 mol}$$.

With this information, you can convert moles to grams by multiplying the moles by the molar mass. Thus, $$\pu{0.806mol} \times \pu{129.839 \frac{g}{mol}}$$ which equals $$\pu{104.68 g}$$. This is your answer.