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A student dissolves 12.48 g of bluestone (CuSO4.5H₂O) in sufficient water to make up 200 mL of solution.

(A) What is the concentration of the solution?

(B) If the student takes 100 mL of this solution, what would be the concentration of the 100 mL
sample?

(C) If the 100 mL sample was then heated strongly to drive off all the water, what mass of copper(11)
sulfate residue would remain?

User Anmol
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1 Answer

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(A) Finding the amount of CuSO4.5H2O dissolved in the specified volume of solution is necessary to determine the concentration of the solution.

Given:

Bluestone (CuSO4.5H2O) mass is 12.48 g.

200 mL is the solution's volume.

Moles per litre, or concentration, is a common unit of measurement. The volume must be changed to litres:

200 mL of solution is equivalent to 200/1000, or 0.2 L.

We divide the number of moles of solute by the litres of solution to obtain the concentration (C):

Molar mass of CuSO4.5H2O is equal to the mass of CuSO4.5H2O.

The formula below can be used to determine the molar mass of CuSO4.5H2O:

Cu's atomic mass is 63.55 g/mol.

S has an atomic mass of 32.07 g/mol.

O has an atomic mass of 16.00 g/mol.

H has an atomic mass of 1.01 g/mol.

CuSO4.5H2O's molar mass is equal to 249.70 g/mol (63.55 + 32.07 + (4 * 16.00) + (5 * (2 * 1.01)) g/mol.

CuSO4.5H2O moles are equal to 12.48 g and 249.70 g/mol.

We can now determine the concentration:

C = Molecular weight of CuSO4.5H2O / litres of solution

Moles of CuSO4.5H2O = 12.48 g / 249.70 g/mol

(B) The concentration of the 100 mL sample, if the student takes 100 mL of the solution, would be determined using the same formula as in part (A), but using the new volume of the sample (0.1 L) rather than 0.2 L.

(C) CuSO4 would be the only residue left after heating the 100 mL sample vigorously to completely evaporate the water. We must use stoichiometry and the molar mass of CuSO4 (minus the water molecules) to get the mass of the residue.

The following formula can be used to determine the molar mass of CuSO4: Molar mass of CuSO4 = 63.55 + 32.07 + (4 * 16.00) = 159.61 g/mol

Stoichiometry enables us to determine that 1 mole of CuSO4.5H2O yields 1 mole of CuSO4. As a result, the moles of CuSO4 and CuSO4.5H2O would be equal.

To determine the residue's mass:

CuSO4 residue mass equals moles of CuSO4.5H2O times the molar mass of CuSO4.

Please let me know the options for parts (B) and (C) so I can assist you.

User NonlinearFruit
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