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Answer is: the freezing point of the solution of sucrose is -0.435°C.
m(H₂O) = 100 g ÷ 1000 g/kg = 0.1 kg.
m(C₁₂H₂₂O₁₁) = 8.0 g. 
n(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).
n(C₁₂H₂₂O₁₁) = 8.0 g ÷ 342.3 g/mol.
n(C₁₂H₂₂O₁₁) = 0.0233 mol.
b(solution) = n(C₁₂H₂₂O₁₁) ÷ m(H₂O).
b(solution) = 0.0233 mol ÷ 0.1 kg.
b(solution) = 0.233 m.
ΔT = b(solution) · Kf(H₂O).
ΔT = 0.233 m · 1.86°C/m.
ΔT = 0.435°C.
Tb = 0°C - 0.435°C = -0.435°C.

The freezing point is the temperature at which the fluid freezes to a solid form. The freezing point of the solution is -0.435 degrees celsius.

What is the freezing point?

The freezing point is the product of the molality, van 't Hoff factor, and the cryoscopic constant. It is given as,

[tex]\rm \Delta T_{F} = K_{F} \times b\times i[/tex]

Given,

Mass of water = 0.1 kg

Mass of sucrose = 8.0 gm

Moles of sucrose are calculated as:

[tex]\begin{aligned}\rm moles &= \dfrac{8.0}{342.3}\\\\&= 0.0233 \;\rm mol \end{aligned}[/tex]

The molality of sucrose is calculated as:

[tex]\begin{aligned}\rm b &= \dfrac{\text{moles of sucrose}}{\text{mass of water}}\\\\&= \dfrac{0.0233 \;\rm mol}{0.1}\\\\&= 0.233 \;\rm m\end{aligned}[/tex]

The freezing point depression is calculated as:

[tex]\begin{aligned}\rm \Delta T &= 0.233 \;\rm m \times 1.86\; ^{\circ} \;\rm C/m\\\\&= 0.435 ^{\circ}\;\rm C\end{aligned}[/tex]

Therefore, the freezing point of a solution is -0.435 degrees celsius.

Learn more about a freezing point here:

https://brainly.com/question/17091798

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