Answer:
a. -7.44 °C
Explanation:
Hello there!
In this case, since the freezing point depression formula is:
[tex]\Delta T_f=-i*m*Kf[/tex]
Thus, since the Van't Hoff's factor is 2 for KCl as it ionizes in K⁺ and Cl⁻, the molarity is 2.0 m (2.0mol/1.0kg) and the freezing point depression constant is 1.86 °C/m, we calculate the freezing point depression as follows:
[tex]\Delta T_f=-2*2.0m*1.86\°C /m\\\\\Delta T_f=-7.44\°C[/tex]
Therefore, the answer is a. -7.44 °C.
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