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Step-by-Step Solution
Step 1: Understand the Given Data
• Total mass of the solution = 1 kg = 1000 g
• Molality of the sucrose solution (m) = 0.75 molal
(This means 0.75 moles of sucrose per 1 kg of water.)
• The solution is cooled to −4°C before freezing
• Cryoscopic constant for water, K_f = 1.86\ \text{K kg mol}^{-1}
Step 2: Express Molality in Terms of Unknown Mass of Water
Let the mass of water in the 1 kg solution be x grams.
Then, the mass of sucrose in the solution = (1000 - x) grams.
The molality definition states:
\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}
For sucrose (molecular mass = 342\ \text{g mol}^{-1} ):
\text{moles of sucrose} = \frac{(1000 - x)}{342}
Mass of water in kg = \frac{x}{1000} .
Hence,
0.75 = \frac{\frac{(1000 - x)}{342}}{\frac{x}{1000}}
Step 3: Solve for the Mass of Water ( x )
Rearranging the above equation:
0.75 = \frac{(1000 - x)}{342} \times \frac{1000}{x}
0.75 \times \frac{x}{1000} = \frac{(1000 - x)}{342}
Multiply both sides appropriately:
0.75 \times x \times 342 = 1000 \times (1000 - x)
256.5\,x = 10^6 - 1000\,x
1256.5\,x = 10^6
x \approx 795.86\ \text{g}
Thus, the mass of water in the original solution is about 795.86 g.
Step 4: Use Freezing Point Depression to Determine Final Mass of Water
The freezing point depression \Delta T_f is given as 4°C (since it is cooled down to −4°C from 0°C).
The relation for freezing point depression is:
\Delta T_f = K_f \times \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}.
Let the new mass of water (after ice separation) be a kg. The moles of sucrose remain the same as in the original solution (since solute does not freeze out):
\text{moles of sucrose} = \frac{(1000 - x)}{342} \approx 0.5969.
Hence,
4 = 1.86 \times \frac{0.5969}{a}.
Solve for a :
a = 1.86 \times \frac{0.5969}{4} \quad \text{(rearranged appropriately)}
or more directly:
4 = 1.86 \times \frac{0.5969}{a} \quad \Rightarrow \quad a = \frac{1.86 \times 0.5969}{4}.
a \approx 0.2775\ \text{kg} = 277.5\ \text{g}.
Step 5: Calculate the Mass of Ice Separated
Initially, the solution had 795.86 g of water. After cooling and solidification, there is only 277.5 g of water in the liquid phase (still part of the solution). The difference will be the mass of ice separated.
\text{Mass of ice separated} = 795.86 - 277.5 = 518.36\ \text{g}.
Rounding to the nearest integer gives \boxed{518\ \text{g}} .
