Question

what mas of water is needed to dissolve 34.8g of copper(II) sulfate in order to prepare a 0.521 m solution?

Answer 1

kimh that could be a trick question cause copper (II) sulfate first apsorbs water and then creates copper (II) sulfate hepta hydrate (CuSO4*7H2O) and then dissolves... so lets say that you have pure copper (II) sulfate which is CuSO4 and you say you have:
\[m(CuSO _{4}) = 34,8 g\]
\[M(CuSO _{4}) = 159,607 g mol ^{-1}\]
and you need to prepare 0,521 M solution.. so therefore:
\[c = n/V ; n = m/M \]
\[n(CuSO _{4}) = m(CuSO _{4})/M(CuSO _{4}) = 34,8g/159,607g mol ^{-1} = 0,2180 mol\]
\[V(H _{2}O) = n(CuSO _{4})/c(CuSO _{4})_{aq} = 0,2180 mol / 0,521 mol dm ^{-3} = 0,4184 dm ^{3}\]
and now if you remember that density of water is approximately 1 g/cm3 or 1 kg/dm3 you have:
\[\rho = m/V \rightarrow m(H _{2}O) = \rho (H _{2}O) \times V(H _{2}O) = 1 kg dm ^{-3} \times 0,4184 dm ^{3}= 0,4184 kg\]

I hope this helps you...