Question

what mass of fe (no3)2 is required to make 800 ml of a 0.75mol/l solution of fe(no3)2?

Answer 1

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.