Problem 76 - A Complete Resource Book in Chemistry for JEE Main

Question

A solution containing \(10 \mathrm{~g}\) per \(\mathrm{dm}^{3}\) of urea (molecular mass \(=60 \mathrm{~g} \mathrm{~mol}^{-1}\) ) is isotonic with a \(5 \%\) solution of a nonvolatile solute. The molecular mass of this non volatile solute is (a) \(350 \mathrm{~g} \mathrm{~mol}^{-1}\) (b) \(200 \mathrm{~g} \mathrm{~mol}^{-1}\) (c) \(250 \mathrm{~g} \mathrm{~mol}^{-\mathrm{t}}\) (d) \(300 \mathrm{~g} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The molecular mass of the non-volatile solute is approximately 300 g/mol.

Step 1: Calculate Molarity of Urea Solution

The urea solution has a concentration of \(10 \text{ g/dm}^3\). Given the molecular mass of urea is \(60 \text{ g/mol}\), the molarity \(M_1\) can be calculated as follows:\[ M_1 = \frac{10 \text{ g/dm}^3}{60 \text{ g/mol}} = \frac{1}{6} \text{ mol/dm}^3 \approx 0.167 \text{ mol/dm}^3 \]

Step 2: Recognize Isotonic Condition

Since the two solutions are isotonic, they exert the same osmotic pressure. Therefore, their molarities can be considered equal for this calculation, as the osmotic pressure depends on molarity for non-volatile solutes.

Step 3: Calculate Molarity of Non-Volatile Solute Solution

The given non-volatile solute solution is a \(5\%\) solution, meaning it contains \(5\text{ g of solute per } 100\text{ mL}\). This is equivalent to \(50\text{ g of solute per }1\text{ dm}^3\). Let the molar mass of this solute be \(M\). The molarity \(M_2\) is:\[ M_2 = \frac{50\text{ g/dm}^3}{M\text{ g/mol}} \]

Step 4: Equate Molarities and Solve for M

By equating the molarity of the isotonic solutions, we have:\[ 0.167 \text{ mol/dm}^3 = \frac{50\text{ g/dm}^3}{M\text{ g/mol}} \]Solving for \(M\):\[ M = \frac{50}{0.167} \approx 299.4 \text{ g/mol} \]

Step 5: Choose the Closest Option

The calculated molecular mass of the non-volatile solute is approximately \(299.4 \text{ g/mol}\). The closest option provided is:(d) \(300 \text{ g/mol}\)

Key Concepts

Molarity

Molarity is a way to express the concentration of a solution. It tells us the number of moles of solute (the substance being dissolved) present in one liter of solution. It is usually denoted by the symbol \( M \). Molarity is calculated using the formula:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
This is important because it allows chemists to accurately create solutions with precise properties, which is especially valuable in reactions. For example, in the given exercise, the molarity of the urea solution was calculated by dividing the mass of urea by its molecular mass, resulting in a molarity \( M_1 \) of approximately \( 0.167 \text{ mol/dm}^3 \).
Understanding molarity is key to predicting how substances in solution will interact, such as determining if solutions are isotonic. Molarity affects processes like diffusion and osmotic pressure, which are crucial in biological and chemical systems.

Osmotic Pressure

Osmotic pressure is a physical property of solutions which arise from the tendency of a solvent to move through a semipermeable membrane into a more concentrated solution. This pressure is directly related to the concentration of solute molecules.
If two solutions are isotonic, it means they have the same osmotic pressure. This occurs when the solutions have equal molar concentrations of solute, as osmotic pressure depends on molarity and temperature. In the provided problem, the urea solution and the non-volatile solute solution are described as isotonic. Therefore, despite having different solute particles, they exert the same osmotic pressure because their molarity is the same.
This principle is vital in biological systems, where cell membranes allow the passage of solvent to balance the osmotic pressure, preventing cell damage from swelling or shrinking.

Molecular Mass Calculation

To determine the molecular mass of an unknown solute, we can use information about its concentration and how it behaves in solution. From the exercise, we know that a 5% solution of a non-volatile solute is isotonic with a known urea solution.
Here is how we go through the calculation:

• Calculate the molarity of the known solution (urea in this case).
• The osmotic pressure, being equal, allows equating the molarity of the two solutions.
• Use the percentage concentration to express how much solute is in a given volume, then equate that to the known molarity.
• This allows solving for the unknown molecular mass, \( M \), giving us \( M \approx 299.4 \text{ g/mol} \).

By following these steps, we can deduce molecular characteristics of unknown solutes, making it a powerful tool in chemistry for identifying or confirming the identity of chemical substances.