Question

If 75 g. of Potassium Chloride (ionic compound) is dissolved in 250 grams of

water, what will be the freezing point of the solution? Kf = 1.86

Answer 1

`T_(f,sol)=-15.0\°C`

Step-by-step explanation:

Hello there!

In this case, for these problem about the colligative property of freezing point depression, it is possible set up the following equation:

`T_(f,sol)-T_(f,water)=-i*m*Kf`

Whereas the van't Hoff's factor, i, is 2 since KCl is ionized in two ions (K+ and Cl-); and the molality (m) of the solution is computed by:

`m=(75g*(1mol)/(74.55g) )/(250g*(1kg)/(1000g) ) \\\\m=4.02mol/kg`

Thus, since the freezing point of water (ice) is 0°C, we obtain the following freezing point of the solution by plugging in:

`T_(f,sol)=-(2)(4.02mol/kg)(1.86\°C/mol*kg)\\\\T_(f,sol)=-15.0\°C`

Best regards!