anonymous
  • anonymous
what mas of water is needed to dissolve 34.8g of copper(II) sulfate in order to prepare a 0.521 m solution?
OCW Scholar - Introduction to Solid State Chemistry
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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...

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