Problem 78 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(0.44 \mathrm{~g}\) of substance dissolved in \(22.2 \mathrm{~g}\) of benzene lowers the freezing point of benzene by \(0.567^{\circ} \mathrm{C}\), then the molecular mass of substance is, (the molal depression constant \(\left.=5.12^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg}\right)\) (a) \(128.4\) (b) \(156.6\) (c) \(178.9\) (d) \(232.4\)

Step-by-Step Solution

Answer

The molecular mass of the substance is 178.9.

Step 1: Understanding the Freezing Point Depression Formula

The freezing point depression of a solution is given by the formula: \[\Delta T_f = i \cdot K_f \cdot m\]where \(\Delta T_f\) is the change in freezing point, \(K_f\) is the molal depression constant, \(m\) is the molality of the solution, and \(i\) is the van't Hoff factor (which is 1 for non-electrolytes). In this case, we have \(\Delta T_f = 0.567^{\circ} \mathrm{C}\) and \(K_f = 5.12^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg}\).

Key Concepts

Molal Depression Constant

The molal depression constant, often represented as \(K_f\), is a key concept in the study of colligative properties of solutions. It is unique to each solvent and reflects how much the freezing point decreases per molal concentration of solute. In essence, \(K_f\) is a measure of the solvent's ability to reduce its freezing point when a solute is added. The units of \(K_f\) are degrees Celsius per molal \( (\text{°C} \cdot \text{mol}^{-1} \cdot \text{kg}) \), meaning for each mole of solute added to one kilogram of solvent, the freezing point decreases by \(K_f\) degrees Celsius.

In the exercise, benzene's molal depression constant is provided as \(5.12 \text{°C} \cdot \text{mol}^{-1} \cdot \text{kg}\). This indicates that for every mole of solute dissolved in one kilogram of benzene, the freezing point will lower by \(5.12\) degrees Celsius. Understanding this value helps in calculating the molality and ultimately finding the molecular mass of the solute.

Molecular Mass Calculation

The calculation of molecular mass from freezing point depression involves several steps, starting with finding the molality \(m\) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent. Using the formula for freezing point depression, \(\Delta T_f = i \cdot K_f \cdot m\), we can solve for molality:

• \(m = \frac{\Delta T_f}{i \cdot K_f}\)
• In our exercise, \(\Delta T_f = 0.567 \text{°C}\) and \(K_f = 5.12 \text{°C} \cdot \text{mol}^{-1} \cdot \text{kg}\)

For non-electrolytes, the van't Hoff factor \(i\) is \(1\), simplifying our calculation. Once we find molality, we can determine the number of moles of solute present. Substituting back into the molecular mass formula, \(\text{Molecular Mass} = \frac{\text{mass of solute}}{\text{moles of solute}}\), allows us to compute the solute's molecular mass.

Finally, plugging the values into the equation yields the molecular mass, in our problem, which helps identify the correct choice among the options provided.

van't Hoff Factor

The van't Hoff factor, symbolized as \(i\), accounts for the number of particles a solute produces in solution. For colligative properties like freezing point depression, it is crucial to understand how \(i\) affects these properties. In many simple calculations, especially involving non-electrolytes like in our exercise, \(i\) equals \(1\) as the solute does not dissociate into ions. For electrolytes, \(i\) would correspond to the number of ions formed.

• If a substance dissociates into two ions, \(i = 2\)
• For non-dissociating solutes, \(i = 1\)
• This distinction is significant because the factor amplifies the effect of solutes in modifying the freezing point

Having \(i = 1\) simplifies our calculations for the molecular mass, as seen in the exercise. It is the same as multiplying by one, which leaves values unchanged but must be verified before performing calculations with other solutes.