Question

A saturated solution prepared at 70∘C contains 32.0 g CuSO4 per 100.0 g solution. A 350 −g sample of this solution is then cooled to 0∘C and CuSO4⋅5H2O crystallizes out. If the concentration of a saturated solution at 0∘C is 12.5 gCuSO4/100 g soln, what mass of CuSO4⋅5H2O would be obtained? [Hint: Note that the solution composition is stated in terms of CuSO4 and that the solid that crystallizes is the hydrate CuSO4⋅5H2O.] Express your answer using two significant figures.

Answer 1

Answer:

Explanation:

 gram of Copper sulphate in 350 g solution is 32 x 3.5 = 112

in term of mol = 112 / 159.5 = 0.7022

after crystallisation, gm of copper sulphate in the solution

= 43.75

in term of mole

43.75/159.5 = 0.2743

mole that got precipitated in hydrated form = 0.7022 - 0.2743 =0.4279

in terms of gm of hydrated copper sulphate = .4279 x mol weight of hydrated copper sulphate = .4279 x 249.5 = 106.76 g .

Answer 2

Answer:

$1.6x10^{2}gCuSO_4\ ^.5H_2O$

Explanation:

Hello,

In this case, the mass of copper (II) sulfate that are available for the 350-g sample is:

$m_{CuSO_4}=0.32\frac{gCuSO_4}{gSample} *350gSample=112gCuSO_4$

And the mass of water:

$m_{H_2O}=350g-112g=238gH_2O$

Now, the dissolved copper (II) sulfate at 0°C is:

$350gSample*0.125\frac{gCuSO_4}{gSample} =43.75gCuSO_4$

Therefore, the undissolved copper (II) sulfate is:

$m_{CuSO_4}^{undissolved}=112-43.75=68.25gCuSO_4$

In addition, 238 g of water are equivalent to 13.5 moles of water, nevertheless, as a pentahydrate is obtained, 5 moles of water correspond to:

$m_{H_2O}^{crystalized}=\frac{5mol*238g}{13.2mol}=90.152g H_2O$

Therefore, the mass of copper (II) sulfate pentahydrate turns out:

$m_{CuSO_4\ ^.5H_2O}=68.25g+90.152g=1.6x10^{2}gCuSO_4\ ^.5H_2O$

Best regards.