Colligative Properties question 9?

The boiling point of water at 735 mmHg is 99.073°C.

What mass of NaCl should be added to 7.31kg water to increase the boiling point to 100.000°C ?

Kb for water = 0.510 K kg mol-1.

Answer:
delta T = 100.000 - 99.073 = 0.927 °C

delta T = kf x m x n

n is the number of ions : NaCl >> Na+ + Cl- 2 ions n=2

m is the molality = moles solute / Kg water

0.927 = 0.510 x m x 2

m = 0.909 = moles NaCl / 7.31 Kg

moles NaCl = 6.64

molar mass NaCl = 58.45 g/mol

6.64 mol x 58.45 g / mol = 388 g NaCl

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Colligative Properties question 9?

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