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

Question

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

its 287.54

anonymous · Answer

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

anonymous · Answer

Hold on there is more

anonymous · Answer

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

anonymous · Answer

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