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








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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) 




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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