A solution containing $62\, g$ ethylene glycol in $250 \,g$ water is cooled to $-10^{\circ}C$. If $K_f$ for water is $1.86\, K\, kg \,mol^{-1}$, the amount of water (in g) separated as ice is :
- 32
- 48
- 16
- 64
The Correct Option is D
Solution and Explanation
To determine the amount of water separated as ice when the solution is cooled to \(-10^{\circ}C\), we use the concept of freezing point depression. The depression in freezing point is given by the formula:
\(\Delta T_f = i \cdot K_f \cdot m\)
where:
- \(\Delta T_f\) is the depression in freezing point.
- \(i\) is the van 't Hoff factor, which is \(1\) for non-electrolyte solutions.
- \(K_f\) is the cryoscopic constant, for water it is \(1.86 \, \text{K kg mol}^{-1}\).
- \(m\) is the molality of the solution.
Step 1: Calculate the molality (m) of ethylene glycol
The molar mass of ethylene glycol (\(\text{C}_2\text{H}_6\text{O}_2\)) is \(62 \, \text{g/mol}\). Therefore, the molality is calculated as follows:
\(m = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)} \times \text{mass of solvent (kg)}}\)
\(m = \frac{62}{62 \times 0.25} = 4\, \text{mol/kg}\)
Step 2: Calculate the freezing point depression
The freezing point depression is calculated using:
\(\Delta T_f = i \cdot K_f \cdot m = 1 \times 1.86 \times 4 = 7.44 \, \text{K}\)
The normal freezing point of water is \(0^{\circ}C\). So, the depressed freezing point is:
\(0^{\circ}C - 7.44^{\circ}C = -7.44^{\circ}C\)
Step 3: Determine how much water needs to freeze to achieve a temperature of \(-10^{\circ}C\)
Since the given temperature is \(-10^{\circ}C\), which is lower than \(-7.44^{\circ}C\), some water must separate as ice to achieve this temperature.
The new molality when water separates as ice can be re-calculated:
Let \(x\) be the mass of water (in kg) that separates as ice:
The new mass of water remaining = \(0.25 - x\) kg.
New molality = \(m' = \frac{62}{62 \times (0.25 - x)}\)
At equilibrium (to maintain \(-10^{\circ}C\)):
\(10 = 1.86 \times \frac{1}{0.25 - x}\)
Solving for \(x\):
\(10 = 1.86 \times \frac{1}{0.25 - x}\)
\(0.25 - x = \frac{1.86}{10}\)
\(0.25 - x = 0.186\)
\(x = 0.25 - 0.186 = 0.064 \, \text{kg} = 64 \, \text{g}\)
Thus, the amount of water separated as ice is 64 g.
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Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.