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In depression of freezing point method c...

In depression of freezing point method camphor is a suitable solvent as its

A

`K_(f)` is high

B

Sublimation is easier

C

Volatility is large

D

Density is low

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The correct Answer is:
To determine why camphor is a suitable solvent in the depression of freezing point method, we can analyze the options provided and understand the underlying principles of colligative properties. ### Step-by-Step Solution: 1. **Understanding Depression of Freezing Point**: - The depression of freezing point (\( \Delta T_f \)) is a colligative property, which means it depends on the number of solute particles in a solution rather than their identity. - The formula for depression of freezing point is given by: \[ \Delta T_f = K_f \times m \] where: - \( \Delta T_f \) = depression in freezing point - \( K_f \) = molal depression constant of the solvent - \( m \) = molality of the solution 2. **Identifying the Importance of \( K_f \)**: - The value of \( K_f \) is crucial because a higher \( K_f \) means a greater depression in freezing point for a given molality of solute. - Therefore, a solvent with a high \( K_f \) will allow for easier determination of molar mass through freezing point depression experiments. 3. **Evaluating the Options**: - **Option 1**: "KF is high" - This is true. Camphor has a high \( K_f \), making it a suitable solvent for this method. - **Option 2**: "Sublimation is easier" - This is incorrect because sublimation does not relate to freezing point depression. - **Option 3**: "Volatility is large" - This is also incorrect. While volatility is a property of the solvent, it does not directly affect the freezing point depression. - **Option 4**: "Density is low" - This is not relevant to freezing point depression. 4. **Conclusion**: - The correct answer is that camphor is a suitable solvent because its \( K_f \) is high, which enhances the ability to measure the depression in freezing point effectively. ### Final Answer: Camphor is a suitable solvent in the depression of freezing point method because its \( K_f \) is high. ---
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