Asolution is made by dissolving 3.8 moles of sodium chloride (nacl) in 185 grams of water. if the molal boiling point constant for water (kb) is 0.51 °c/m, what would be the boiling point of this solution? show all of the work needed to solve this problem.

The boiling point of the solution can be calculated by the equation of the colligative property named boilng point elevation.

The formula for the boiling point elevation is:

ΔTb = i * Kb * m

Where:

i = van't Hoof factor
Kb = molal boiling point constant
m = molality.

Solution:

1) i is 2 because the NaCl is ionice and dissociates into two ions.

2) Kb is given = 0.51 °C/m

3) m = moles of solute / kg of solvent

moles of solute is given = 3.8

kg of solvent = 185 g * 1 kg/1000g = 0.185 kg

m = 3.8 moles / 0.185 kg = 20.54 m

4) ΔTb = 2 * 0.51°C/m * 20.54m = 20.9 °C

5) The boiling point is the normal boiling point of water (100°C) plus the increase:

Tb = 100°C + 20.9°C = 120.9°C

120.9°C.

The answer is b metal denting

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