Question

After a long ski day a salt bath of NaCl is mixed so that the melting point is - 2 °C in Lake Tahoe at an altitude of 1900 m. What is the boiling point of this solution? The atmospheric pressure in Lake Tahoe is 0.80 atm on average.

Answer 1

The depression in freezing point (ΔT_f ) = i * K_f* m

ΔT_f = Frrezing point depression of teh lake water = New freezing point - ( Normal freezing point of water) = - 2 - 0 = - 2 ⁰C

Where i = Vanthoff's factor = it depends on the numbers of ions produced = 2 ( for NaCl)

Kf = freezing point depression constant of water = 1.86 ⁰C/ m

m = Molality of NaCl in solution = X (m)

So, 2 ⁰C = 2 * 1.86 ⁰C/m * X (m)

Or, X = 0.538

The elevation in boiling point (ΔT_b ) = i * K_f* m

Here, Kb = boiling point elevation constant of water = 0.512 ⁰C/ m

ΔT_b = 2 * 0.512 ⁰C/ m * 0.538 (m) = 0.55 ⁰C

The boiling point of this solution = 55 + 100 = 100.55 ⁰C