Question

Given the formula for depression of freezing point: \[ \Delta T_f = K_f \cdot m \] where \( \Delta T_f \) is the depression of freezing point, \( K_f \) is the freezing point depression constant, and \( m \) is the molality, calculate the value of \( m \).

• A. \( m = \frac{\Delta T_f}{K_f} \)
• B. \( m = \frac{K_f}{\Delta T_f} \)
• C. \( m = K_f \cdot \Delta T_f \)
• D. \( m = \frac{\Delta T_f}{K_f^2} \)

Hint:

When using the depression of freezing point formula, rearrange the equation to solve for the desired variable. In this case, the molality \( m \) can be calculated by dividing the depression of freezing point \( \Delta T_f \) by the freezing point depression constant \( K_f \).

Answer 1

The Correct Option is A

The depression of freezing point is given by the equation: \[ \Delta T_f = K_f \cdot m \] Where: - \( \Delta T_f \) is the depression in freezing point, - \( K_f \) is the freezing point depression constant, and - \( m \) is the molality of the solution.
Step 1: Solve for molality To find the molality \( m \), we can rearrange the formula: \[ m = \frac{\Delta T_f}{K_f} \] This equation shows that molality is directly proportional to the depression in freezing point and inversely proportional to the freezing point depression constant. Thus, the correct answer is \( \boxed{m = \frac{\Delta T_f}{K_f}} \).