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)

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    $\begingroup$ This is definitely a homework question. You must need to show your effort to solve it. $\endgroup$ – Mathew Mahindaratne Mar 19 '19 at 2:03
  • $\begingroup$ 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... $\endgroup$ – Nam Huy Mar 19 '19 at 2:05
  • $\begingroup$ 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. $\endgroup$ – Mathew Mahindaratne Mar 19 '19 at 2:12
  • $\begingroup$ I believe it is .807, what do I do after this? $\endgroup$ – Nam Huy Mar 19 '19 at 2:18
  • $\begingroup$ Correct answer is $\pu{0.806 mol}$. So, do you remember, $g= \pu{mol} \times \text{molar mass}$? $\endgroup$ – Mathew Mahindaratne Mar 19 '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.


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