1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to \(-4^\circ C\) before freezing. The amount of ice (in g) that will be separated out is _________. (Nearest integer)
[ Given : \(K_f(H_2O) = 1.86 \, \text{K kg mol}^{-1}\) ]
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Be careful with the definition of "1 kg of solution". It is not the same as "solution prepared with 1 kg of solvent". Use ratios to find the exact mass of components.
Step 1: Understanding the Concept:
When a solution is cooled below its initial freezing point, ice (solvent) separates out, making the remaining solution more concentrated until its freezing point matches the current temperature. Step 2: Detailed Explanation:
1. Calculate mass of solute and solvent in the initial 1 kg solution:
A 0.75 molal solution has 0.75 moles of sucrose in 1000 g of water.
Molar mass of sucrose (\(C_{12}H_{22}O_{11}\)) \(= 342 \, \text{g/mol}\).
Mass of sucrose in standard 0.75 molal solution \(= 0.75 \times 342 = 256.5 \, \text{g}\).
Total mass of standard solution \(= 1000 + 256.5 = 1256.5 \, \text{g}\).
In our 1000 g (1 kg) solution:
Mass of sucrose (\(W_2\)) \(= \frac{256.5}{1256.5} \times 1000 \approx 204.14 \, \text{g}\).
Mass of water (\(W_1\)) \(= 1000 - 204.14 = 795.86 \, \text{g}\).
Moles of sucrose \(= \frac{204.14}{342} = 0.5969 \, \text{mol}\).
2. Calculate required mass of water at \(-4^\circ C\):
\(\Delta T_f = 4 \, K\).
\(\Delta T_f = K_f \times m \implies 4 = 1.86 \times \frac{0.5969}{W_{\text{water, final (kg)}}}\).
\[ W_{\text{water, final (kg)}} = \frac{1.86 \times 0.5969}{4} = 0.27756 \, \text{kg} = 277.56 \, \text{g} \]
3. Calculate amount of ice separated:
Ice separated \(= \text{Initial water} - \text{Remaining water}\)
\[ \text{Ice} = 795.86 \, \text{g} - 277.56 \, \text{g} = 518.3 \, \text{g} \] Step 3: Final Answer:
The amount of ice separated is approximately 518 g.
