Question

For each of the -3.3 trials, and in the same sequence in which you entered the freezing points, enter your calculated values for the freezing point depression constant of water. Be sure to use the average value for the freezing point of pure water and the value for the molality of the solution that you entered earlier. Report your values to 3 significant figures.

Answer 1

Answer:

The morality of the solution is calculated as 0.859 m. We are required to determine the freezing point depression constant of pure water. The freezing point depression of the solution is given as

$T_{f} = T_{p} -T_{s} = K_{f}$ * (morality of solution)

$T_{p}$ and $T_{s}$ are the freezing points of the pure solvent (water, 0°C) and$K_{f}$ = freezing point depression constant of water. Therefore,

$T_{f} = K_{f}$ *(0.859 m)

=====> (0°C) – (3.00°C) = $K_{f} *(0.859 m)$

=====> -3.00°C = $K_{f} *(0.859 m)$

Ignore the negative sign (since $K_{f}$ is positive) and get

$K_{f}$ = (3.00°C) / (0.859 m) = 3.492°C/m

The freezing point depression constant of the solvent is 3.5°C kg/mole  

3.5 K.kg/mole (temperature differences are the same in Celsius and Kelvin scales).