Question

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

Answer 1

Answer:

$T_S=-1.09\°C$

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:

$(T_S-0\°C)=-3*0.195m*1.86\frac{\°C}{m}\\\\T_S=-1.09\°C$

Best regards!

Answer 2

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