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The molar mass of a nonvolatile solute is 176.0 g/mol. Compute the boiling point of a solution containing 68.0 g of the solute and 750.0 g of water when the barometric pressure is such that pure water boils at 99.825 °C. (\(K_b = 0.52 \, °C/m\))

Choose one:
A. 99.8 °C
B. 100.1 °C
C. 102.4 °C
D. 98.5 °C
E. 101.2 °C

Answer :

Final answer:

The boiling point of a solution with 68.0 g of a nonvolatile solute with a molar mass of 176.0 g/mol in 750.0 g of water, when water boils at 99.825°C, is calculated to be 100.1°C, which is answer choice (b).

Explanation:

To compute the boiling point of a solution containing a nonvolatile solute with a given molal boiling point elevation constant, we first find the molality of the solution and then apply the boiling point elevation formula.

The molar mass of the nonvolatile solute is 176.0 g/mol, and we have 68.0 g of this solute dissolved in 750.0 g of water. We can calculate the molality (m) by dividing the moles of solute by the kilograms of solvent:

  • Number of moles of solute = (68.0 g) / (176.0 g/mol) = 0.386 mol
  • Mass of solvent in kg = 750.0 g = 0.750 kg
  • Molality (m) = moles of solute / mass of solvent in kg = 0.386 mol / 0.750 kg = 0.515 m

Now, we apply the boiling point elevation formula ΔT = Kb × m, where ΔT is the boiling point elevation and Kb is the molal boiling-point elevation constant, which is provided as 0.52 °C/m for the situation where pure water boils at 99.825 °C:

  • ΔT = (0.52 °C/m) × (0.515 m) = 0.268 °C

Therefore, the new boiling point of the solution is:

  • Boiling point of solution = boiling point of pure solvent + ΔT = 99.825 °C + 0.268 °C = 100.093 °C

After rounding to one decimal place, the boiling point of the solution is 100.1°C, making the correct answer choice (b) 100.1°C.

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