At what temperature will 0.100 molal (M) NaCl(aq) boil? Kb= 0.51 C kg mol^-1

1 answer:

7 0

The increase of the boling point of a solution is a colligative property.

The formula for the increase of the normal boiling point of water is:

ΔTb = Kb * m

Where m is the molallity of the solution and Kb is the molal boiling constant in °C/mol.

ΔTb = 0.51 °C / m * 0.100 m = 0.051 °C.

So, the new boiling temperature is Tb = 100°C + 0.051°C = 100.051 °C.

Answer: 100.051 °C

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