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Calculate the boiling point of a 4.5 m solution of Na2SO4 in water, assuming 100% dissociation. Given [tex]K_p(H_2O) = 0.52 \, \text{°C/m}[/tex].

A. 97.7 °C
B. 102.3 °C
C. 104.7 °C

Answer :

Final answer:

The boiling point of a 4.5 m solution of Na2SO4 in water is approximately 102.34 °C.

Explanation:

To calculate the boiling point of a solution, we need to use the formula:

ΔTb = Kbm

Where:

  • ΔTb is the boiling point elevation
  • Kb is the molal boiling point constant of the solvent
  • m is the molality of the solution

In this case, we have a 4.5 m (molal) solution of Na2SO4 in water. The molal boiling point constant of water is given as 0.52 °C/m.

Substituting the values into the formula:

ΔTb = (0.52 °C/m) * (4.5 m)

Simplifying the equation:

ΔTb = 2.34 °C

To calculate the boiling point of the solution, we need to add the boiling point elevation to the boiling point of pure water. The boiling point of pure water is 100 °C.

Boiling point of solution = 100 °C + 2.34 °C = 102.34 °C

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