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).

3 0

3 years ago

K decreased by the quotient of 2 and 9

dem82 [27]

The answer is k-2/9 because it says that k is decreased which means u substrate and then it's says the quotient of 2 and 9 which means that u divide so then u get 2/9

7 0

3 years ago