Answer to Question #154981 in General Chemistry for A

Question

the chemist used a solution of iron nitrate, Fe(NO₃)₃ with a concentration of 0.010 mol l^–1. If they were preparing one litre of this solution, what mass of iron nitrate should they weigh out?

For your calculations use the following relative atomic mass (RAM) values:

RAM (Fe) = 56

RAM (N) = 14

RAM (O) = 16

Answer 1

compound = $Fe(NO_3)_2$

molar mass of compound = Fe + 2N + 6O = 56 + 2(14) + 6(16) = 180g/mol

concentration = 0.010 mol L^-1

If the volume of the solution is 1L, the amount of Iron Nitrate = c × V = 0.010 × 1 = 0.010 mol

Since 1 mole of Iron Nitrate = 180g

0.010 mol = x

x = $\dfrac{0.010×180}{1}$ = 1.80g

Therefore, If they were preparing one litre of the solution, 1.80g of iron nitrate should be weighed out.