Answer to Question #91121 in Organic Chemistry for Sam

Question #91121
what mass of fe (no3)2 is required to make 800 ml of a 0.75mol/l solution of fe(no3)2?
1
Expert's answer
2019-06-24T04:20:59-0400

According to the formula of molar concentration:


"C_M=\\frac {n(Fe(NO_3)_2)} {V_{solution}}= \\frac {m(Fe(NO_3)_2)} {M(Fe(NO_3)_2) V_{solution}};"

Now we can transform equation and substitute known values with numbers:


"m(Fe(NO_3)_2)=C_M*M(Fe(NO_3)_2)*V_{solution};""m(Fe(NO_3)_2)=0.75 mol\/L * 180 g\/mol * 0.8L= 108g."

Answer: we need 108 g of iron (II) nitrate to make 800ml of 0.75M solution.


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