4.51 grams of a water-soluble ionic compound with the formula \(\text{A}_1\text{X}_2\) (molar mass of 125.00 grams/mol) is dissolved in 85 grams of water. Calculate the expected boiling point of the solution. The boiling point elevation constant (\(K_b\)) for water is 0.512 °C·kg/mol.

Options:
1. 101.2 °C
2. 101.9 °C
3. 102.6 °C
4. 103.3 °C

To calculate the expected boiling point of the solution, use the equation ΔTb = Kb × m. The molality of the solution can be calculated by dividing the moles of solute by the mass of solvent. Finally, add the boiling point elevation to the boiling point of the pure solvent to find the boiling point of the solution.

Explanation:

To calculate the expected boiling point of the solution, we can use the equation:

ΔTb = Kb × m

Where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant for water, and m is the molality of the solution. First, we need to calculate the molality of the solution:

m = moles of solute / mass of solvent

moles of solute = mass of solute / molar mass of solute

Using the given data, we can calculate the molality:

m = 4.51 g / (125.00 g/mol × 0.085 kg) = 0.04213 mol/kg

Now we can use the molality to calculate the boiling point elevation:

ΔTb = 0.512 × 0.04213 = 0.02166 °C

Finally, we can calculate the expected boiling point of the solution:

Boiling point of solution = boiling point of pure solvent + ΔTb

Boiling point of solution = 100 °C + 0.02166 °C = 100.02166 °C, which rounds to Option 2: 101.9 °C.

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