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Question

$\pu{1 kg}$ of a solution of cane sugar is cooled and maintained at $\pu{-4 ^\circ C}$. How much ice will separate out if molality of solution is $0.75$?

$K_\mathrm{f} \ \text{(water)} = 1.86\ \mathrm{K\ kg/mol}$

Solution given:

$\Delta T_\mathrm{f} = 0 - (-4)=4$

$i = 1$ for sucrose.

Therefore, since $\Delta T_\mathrm{f}$ is defined as

$$\Delta T_\mathrm{f}= K_\mathrm{f}\cdot i \cdot \frac {\text{amount of sucrose in } \pu{mol}}{\text{weight of water left in }\pu{kg}}$$

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My doubts

$\Delta T_\mathrm{f}$ is the absolute value of the difference of freezing point of pure solvent and the solution with the non volatile solute. Here, the freezing point of the solution is not $\pu{-4 ^\circ C}$. So how can we say $\Delta T_\mathrm{f}$ is equal to $4$?

Why does the molality term contain amount of remaining water in the denominator? In the original equation shouldn't it be the original molality here?

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    $\begingroup$ This is because we say the temperature is fixed at -4 when actual freezing point is zero.. $\endgroup$ Jul 28 '20 at 9:52
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    $\begingroup$ Go on, plug the original molality and find the freezing point of the original soultion (which is not $-4^\circ\rm C$, that's true). Think about what happens next. $\endgroup$ Jul 28 '20 at 10:06
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The question: $\pu{1kg}$ of a solution of cane sugar is cooled and maintained at $\pu{-4 ^\circ C}$. How much ice will separate out if molality of solution is $0.75$?

Given: $K_\mathrm{f} \text{ (Water)}= \pu{1.86 K kg/mol}$

The solution is given assuming you understand the concept. It seems you did not completely understand the colligative properties of solutions. Thus I'm giving some clues you how to approach this problem.

The textbook equation you are going to use is:

$$\Delta T_\mathrm{f}= K_\mathrm{f}\times i \times \text{molality of the solution} = K_\mathrm{f}\times i \times \frac {\text{amount of sucrose in }\pu{mol}}{\text{weight of solvent in }\pu{kg}} \tag 1$$

where $i = 1$ for sucrose (sucrose does not ionize in water).

The solution is kept at $\pu{-4 ^\circ C}$. The freezing point of pure water is $\pu{0 ^\circ C}$ ($\therefore \ \Delta T = 0 - (-4)=4$). First what you should do is find the freezing point of given solution using equation $(1)$. If $\Delta T_\mathrm{f} \gt 4$, the game is over. That means, this solution does not freeze at $\pu{-4 ^\circ C}$. If $\Delta T_\mathrm{f} \lt 4$, the game is on. That means, some water will freeze out until the final molality reached the molality of solution, which freeze at $\pu{-4 ^\circ C}$.

For given solution (no units was used):

$$\Delta T_\mathrm{f}= K_\mathrm{f}\times i \times m = 1.86 \times 1 \times 0.75 = \pu{1.395} $$

Thus, this solution freeze at $\pu{-1.395 ^\circ C}$.

Now, you have to find out at what molality of a solution would be just about to freeze at $\pu{-4 ^\circ C}$. That means, $\Delta T_\mathrm{f}= 4$ (still, $i = 1$):

$$\Delta T_\mathrm{f}= K_\mathrm{f}\times i \times m \ \Rightarrow \ m= \frac{\Delta T_\mathrm{f}}{K_\mathrm{f}} = \frac{4}{1.86} = 2.15$$

This means, water in original mixture freeze out until the molarity of solution becomes $\pu{2.15 mol/kg}$.

Put it this way:

  • The original solution has $\pu{0.75 mol}$ of sucrose in $\pu{1 kg}$ of water. The amount of sucrose would not freeze.
  • Thus, how much water would remain with $\pu{0.75 mol}$ of sucrose to keep molality at 2.15?

I'm sure it is given in your solution manual. Regardless, I'll show you the calculations. Suppose the weight of water is $W \ \pu{kg}$ (ignore the units in equation}:

$$ \frac{0.75}{W} = 2.15 \ \Rightarrow \ W = \frac{0.75}{2.15} = 0.349$$

Thus, the weight of water in final solution after freezing at $\pu{-4 ^\circ C}$ is $\pu{0.349 kg}$ (the amount of sucrose is still $\pu{0.75 mol}$).

However, this calculation is in significant deviation from actual amount since the given amount of original solution is $\pu{1.0 kg}$. That means it does not contain $\pu{0.75 mol}$ of sucrose. Now, molecular weight of sucrose is $\pu{342.3 g/mol}$. If amount of sucrose in $\pu{1.0 kg}$ of original solution is $x$, then:

$$\frac{\frac{x}{342.3}}{(1000-x)\times 10^{-3}} =0.75 \ \Rightarrow \ 1.257x = 0.75 \times 342.3 \ \Rightarrow \ x = \frac{256.725}{1.257} = \pu{204.2 g}$$

Therefore, the amount of water in that solution is $\pu{1000-204.2) g} = \pu{795.8 g}$ or $\pu{0.796 kg}$. Thus, amount of sucrose in the of original solution is: $\frac{\pu{204.2 g}}{\pu{342.3 g/mol}} = \pu{0.597 mol}$.

Using these values, we can recalculate the water amount in final solution. Suppose the weight of water in final solution is $W_f \ \pu{kg}$:

$$ \frac{0.597}{W_f} = 2.15 \ \Rightarrow \ W_f = \frac{0.597}{2.15} = 0.278$$

Thus, the weight of water in final solution after freezing at $\pu{-4 ^\circ C}$ is $\pu{0.278 kg}$ (the amount of sucrose is $\pu{0.597 mol}$ making it $\pu{2.15 molal}$).

Thus, the amount of water freeze out from the original $\pu{1.0 kg}$ of solution is:

$$\pu{(0.796-0.278) kg} =\pu{0.518 kg}$$

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