Problem 8 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\(5 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) was dissolved in \(x \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). The change in freezing point was found to be \(3.82^{\circ} \mathrm{C}\). If \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is \(81.5 \%\) ionised, the value of \(x\) \(\left(K_{f}\right.\) for water \(\left.=1.86^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}\right)\) is approximately : (molar mass of \(\mathrm{S}=32 \mathrm{~g} \mathrm{~mol}^{-1}\) and that of \(\mathrm{Na}=23 \mathrm{~g} \mathrm{~mol}^{-1}\) ) (a) \(15 \mathrm{~g}\) (b) \(25 \mathrm{~g}\) (c) \(45 \mathrm{~g}\) (d) \(65 \mathrm{~g}\)

Step-by-Step Solution

Answer

The value of \(x\) is approximately 45 g.

Step 1: Calculate the Molar Mass of Na2SO4

First, we need to determine the molar mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). This compound consists of 2 sodium (Na) atoms, one sulfur (S) atom, and 4 oxygen (O) atoms. The atomic masses are: \(\mathrm{Na} = 23 \, \mathrm{g/mol}\), \(\mathrm{S} = 32 \, \mathrm{g/mol}\), and \(\mathrm{O} = 16 \, \mathrm{g/mol}\). Thus, the molar mass is calculated as:\[ 2 \times 23 + 32 + 4 \times 16 = 142 \, \mathrm{g/mol} \]

Step 2: Calculate the Number of Moles of Na2SO4

We have 5 grams of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). Using its molar mass, we calculate the number of moles:\[ \text{Number of moles} = \frac{5 \, \mathrm{g}}{142 \, \mathrm{g/mol}} \approx 0.0352 \, \mathrm{moles} \]

Step 3: Calculate the Van't Hoff Factor (i)

The ionization of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) produces 2 sodium ions and one sulfate ion. If it's 100% ionized, \(i = 3\). Given 81.5% ionization, the effective \(i\) is:\[ i = 1 + 2 \times 0.815 + 1 \times 0.815 = 2.63 \]

Step 4: Use the Freezing Point Depression Formula

The formula for freezing point depression is:\[ \Delta T_f = i \times K_f \times m \]Where \(\Delta T_f = 3.82^{\circ} \mathrm{C}\), \(K_f = 1.86^{\circ} \mathrm{C} \, \mathrm{kg/mol}\), and \(m\) is the molality (moles of solute per kg of solvent). Rearranging the formula for molality gives:\[ m = \frac{\Delta T_f}{i \times K_f} \approx \frac{3.82}{2.63 \times 1.86} \approx 0.781 \, \mathrm{mol/kg} \]

Step 5: Calculate the Mass of Water (x)

The molality formula is:\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]From Step 2, we have \(0.0352 \, \mathrm{moles}\). Solving for the mass of the solvent:\[ 0.781 = \frac{0.0352}{x/1000} \]Solving this, we get:\[ x = \frac{0.0352}{0.781} \times 1000 \approx 45 \, \mathrm{g} \]

Step 6: Conclusion

The calculated mass of water needed, \(x\), to achieve the given freezing point depression at the specified ionization level is approximately \(45 \, \mathrm{g}\).

Key Concepts

Molar Mass Calculation

In order to solve problems involving solutions, it is crucial to first determine the molar mass of the compound involved. For \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), we start by identifying the elements and their respective number of atoms in the molecule:

• 2 sodium (Na) atoms
• 1 sulfur (S) atom
• 4 oxygen (O) atoms

Then, we use the atomic masses: \(\mathrm{Na} = 23 \, \mathrm{g/mol}\), \(\mathrm{S} = 32 \, \mathrm{g/mol}\), and \(\mathrm{O} = 16 \, \mathrm{g/mol}\). By plugging these into the calculation formula:\[ \text{Molar Mass} = 2 \times 23 + 32 + 4 \times 16 = 142 \, \mathrm{g/mol} \]This calculation ensures that we know the number of grams per mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), which is critical for converting between mass and moles, a key step in solving many chemistry problems. Remembering this step will be helpful in handling quantitative questions in other exercises as well.

Van't Hoff Factor

The Van't Hoff factor, denoted as \(i\), describes how many particles a compound breaks into when dissolved in a solution. For ionic compounds like \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), which dissociates into 2 \(\mathrm{Na}^+\) ions and 1 \(\mathrm{SO}_{4}^{2-}\) ion, the theoretical \(i\) value is initially 3 (i.e., 2 + 1). That's when the compound is fully ionized.
However, ionization is not always complete. If the ionization is only 81.5%, the effective Van’t Hoff factor needs adjustment: \[ i = 1 + 2 \times 0.815 + 1 \times 0.815 = 2.63 \]Here, the calculation uses 1 molecule that doesn't dissociate completely, and you add the fractions of dissociated ions. This factor better reflects the real behavior of the solution, resulting in more accurate calculations for properties like freezing point depression.

Ionization Percentage

Ionization percentage tells us how much of a substance dissociates into ions in a solution. It’s crucial in determining the solution’s actual behavior. For our compound \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), the problem mentions an ionization of 81.5%, signaling that not all molecules break apart: only 81.5% do.
This percentage affects the calculations for properties such as the Van't Hoff factor and plays a key role in real-life applications like calculating the required mass or concentration for a given reaction. In mathematical terms, it adjusts how we approach our solute's impact on the solvent, helping us predict results like freezing point or boiling point changes accurately.

Molality

Molality is a measure of the concentration of a solution expressed as moles of solute per kilogram of solvent. It differs from molarity, which is moles of solute per liter of solution. In problems like freezing point depression, molality is more useful because it doesn't change with temperature.
To find molality, use the expression:\[ m = \frac{\Delta T_{f}}{i \times K_{f}} \]where \(\Delta T_{f}\) is the change in freezing point, \(i\) is the Van’t Hoff factor, and \(K_{f}\) is the cryoscopic constant.In our exercise, calculated molality was about \(0.781 \, \mathrm{mol/kg}\), obtained from the known freezing point change and using the adjusted Van’t Hoff factor. Molality helped us transition to calculating how much water (the solvent) was needed, emphasizing its importance in calculations involving temperature changes.