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AlexFokin [52]
3 years ago
7

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

Chemistry
2 answers:
Elina [12.6K]3 years ago
4 0

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!

Kay [80]3 years ago
4 0

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

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