High School

Calculate the boiling point of a 1.5 m aqueous solution of NaCl. The boiling point constant of water is 0.512 °C/m. Remember, m is a unit for molality.

a. 98.5 °C
b. 100.0 °C
c. 100.8 °C
d. 101.5 °C

Answer :

To determine boiling point, we have the equation: ΔT = mK; where ΔT refers to the boiling-point elevation, m refers to molality which is in the unit mol/kg, and K is the molal boiling-point elevation constant.

In this problem, water is the solvent and NaCl is the solute. Note that 100°C is the boiling point of water. We substitute the equation:
ΔT = (1.5 m)(0.512 °C/m)
ΔT = 0.768°C
This is the change in temperature. To find the boiling point of the whole solution, we add the boiling point of the solvent:
100°C + 0.768°C = 100.768°C ≈ 100.8°C

The answer is C.

Final answer:

To calculate the boiling point of a solution, we use the equation ΔTb = Kb × m. The boiling point elevation is determined by multiplying the boiling point constant of the solvent by the molality of the solution. The boiling point of the 1.5 m NaCl solution is approximately 100.8 °C.

Explanation:

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

ΔTb = Kb × m

where ΔTb represents the boiling point elevation, Kb is the boiling point constant of the solvent, and m is the molality of the solution.

In this case, the molality of the solution is given as 1.5 m, and the boiling point constant of water is 0.512 °C/m. Plugging these values into the equation:

ΔTb = (0.512 °C/m) × (1.5 m) = 0.768 °C

The boiling point elevation is the difference between the boiling point of the solution and the boiling point of the pure solvent. Therefore, the boiling point of the solution would be the boiling point of water (100°C) plus the boiling point elevation:

Boiling point of solution = 100°C + 0.768 °C = 100.768 °C

Rounding to one decimal place, the boiling point of the 1.5 m NaCl solution is approximately 100.8 °C.

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