Question

What is the freezing point in °C) of a 0.195 m

aqueous solution of K2S?
Enter your rounded answer with
3 decimal places.
K for water = 1.86 °C/m

Answer 1

The correct answer is -1.088. (Don't forget the negative sign).

Answer 2

`T_S=-1.09\°C`

Step-by-step explanation:

Hello!

In this case, since the freezing point depression for a solution is computed via:

`(T_S-T_W)=-imK_f`

Whereas TW is the freezing temperature of water, TS that of the solution, i the van't Hoff's factor (3 for K2S as it ionizes properly), m the molality of the solution and Kf the freezing point constant of water. Thus, we plug in to obtain:

`(T_S-0\°C)=-3*0.195m*1.86(\°C)/(m)\\\\T_S=-1.09\°C`

Best regards!