# What is the freezing point of a solution containing 6.5 % KCl by mass (in water).?

May 12, 2018

#### Answer:

Well what is the molal cryoscopic constant for water?

#### Explanation:

This site says $1.853 \cdot K \cdot k g \cdot m o {l}^{-} 1$

And so we must work out the molality of a 6.5% solution....we take a mass of $1 \cdot k g$ of solution....

$\text{Molality"="Moles of solute"/"Kilograms of solvent} = \frac{\frac{65.0 \cdot g}{74.55 \cdot g \cdot m o {l}^{-} 1}}{\left(1000 \cdot g - 65.0 \cdot g\right) \times {10}^{-} 3 \cdot k g \cdot {g}^{-} 1} = 0.933 \cdot m o l \cdot k {g}^{-} 1$...

And of course there is a catch. Potassium chloride undergoes chemical REACTION in water to give two equivs of ions...

$K C l \left(s\right) \stackrel{{H}_{2} O}{\rightarrow} {K}^{+} + C {l}^{-}$

...where each ion is aquated by several water molecules...

And so there are TWO equiv of ion in solution per equiv of salt...

Finally, we get....

$0.933 \cdot m o l \cdot k {g}^{-} 1 \times 2 \times 1.853 \cdot K \cdot k g \cdot m o {l}^{-} 1 = 3.46 \cdot K$

And thus $\text{fusion point"_"solution} = {\left(0.0 - 3.46\right)}^{\circ} C = - 3.46 {\cdot}^{\circ} C$