Question

Calculate the freezing point of a solution containing 5. 0 grams of kcl and 550. 0 grams of water. the molal-freezing-point-depression constant (kf) for water is 1. 86°c/m.

Answer 1

The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

Using the equation,

Δ$T_{f}$ = i$K_{f}$m

where:

Δ$T_{f}$ = change in freezing point (unknown)

i = Van't Hoff factor

$K_{f}$ = freezing point depression constant

m = molal concentration of the solution

Molality is expressed as the number of moles of the solute per kilogram of the solvent.

Molal concentration is as follows;

MM KCl = 74.55 g/mol

molal concentration = $\frac{5.0g*\frac{1mol}{74.55g} }{550.0g*\frac{1kg}{1000g} }$

molal concentration = 0.1219m

Now, putting in the values to the equtaion Δ$T_{f}$ = i$K_{f}$m we get,

Δ$T_{f}$ = 2 × 1.86 × 0.1219

Δ$T_{f}$ = 0.4536°C

So, Δ$T_{f}$ of solution is,

Δ$T_{f_{solution} }$ = 0.00°C - 0.45°C

Δ$T_{f_{solution} }$ =  - 0.45°C

Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is  - 0.45°C

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