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Chemistry 19 Online
OpenStudy (anonymous):

how do I prepare a 0.1 mol solution of znSO4.7H2O which needs to be 100cm3 using the solid znSO4.7H2O and water?

OpenStudy (anonymous):

Ok first you are going to have to find the molecular weight of znso4.7h2o

OpenStudy (anonymous):

After you find that i will do the factor label for you

OpenStudy (anonymous):

its 287.54

OpenStudy (anonymous):

alright first thing i would figure out is how many grams .1 mole of zns04.7h20 is \[.1 mole zn ( \frac{287.54 grams zn}{1 mole zn} )=28.754 grams zn\]

OpenStudy (anonymous):

Hold on there is more

OpenStudy (anonymous):

\[100 cm^3 ( \frac{1ml}{1cm^3} )( \frac{1 mole}{1000ml} ) (\frac{287.54 g}{1 mole})=28.754g\]

OpenStudy (anonymous):

So it looks like you are going to have to dillute 28.754g of znso4.7h2o in 100ml of h20

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