Asaturated 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.

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 .

Explanation:

Hello,

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

And the mass of water:

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

Therefore, the undissolved copper (II) sulfate is:

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:

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

Best regards.

what is the question?

b) nuclear ive learned that in a pedtrodic table