Answer:
- 0.454 °C
Explanation:
From the given information:
The mass of KCl = 10 grams
The molar mass of KCl = 74.55 g/mol
The molality can be calculated as:
[tex]Molality \ m = \dfrac{mass \ of \ KCl \times 1000}{molar \ mass \ of \ KCl \times 1100.0}[/tex]
[tex]Molality \ m = \dfrac{10 \ g \times 1000}{74.55 \ g/mol \times 1100.0}[/tex]
Molality m = 0.1219 M
For freezing point depression;
[tex]\Delta T_R = i \times k_f \times m[/tex]
The van't Hoff's Factor (i) for KCl = 2
The molar depression of freezing point constant [tex]k_f[/tex] = 1.86° C/m
∴
[tex]\Delta T_R = 2 \times 1.86 \times 0.1219[/tex]
[tex]\mathbf{\Delta T_R = 0.454^0\ C }[/tex]
However, the freezing point of water is known to be = 0°C
Therefore, the freezing point of the solution
= 0 °C - 0.454 °C
= - 0.454 °C
