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) \(250 \mathrm{~g} \mathrm{~mol}^{-1}\) (d) \(300 \mathrm{~g} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The molecular mass of the non-volatile solute is 300 g/mol (option d).

Step 1: Understanding Isotonic Solutions

Isotonic solutions have the same osmotic pressure. Since osmotic pressure (c0) depends on the molarity and temperature of the solution, for two solutions to be isotonic, \[ C_1 \cdot i_1 = C_2 \cdot i_2 \]where \(C\) represents the concentration and \(i\) is the van 't Hoff factor. Since both solutions are nonvolatile and not dissociating, \(i = 1\) for both.

Step 2: Calculating Molarity of Urea Solution

The urea solution has 10 grams per liter. Urea's molar mass is 60 g/mol. The molarity (\[M_1\]) of the solution is given by: \[ M_1 = \frac{10 \, \text{g/dm}^3}{60 \, \text{g/mol}} = \frac{1}{6} \text{mol/dm}^3 \]

Step 3: Finding Molarity of Non-Volatile Solution

Assume the molecular mass of the non-volatile solute is \(M_2\). A 5% solution means 5 grams of solute in 100 mL, equivalent to 50 grams per liter. Therefore, the molarity (\[M_2\]) of the non-volatile solution is: \[ M_2 = \frac{50 \, \text{g/dm}^3}{M_2} \]

Step 4: Setting Up Isotonic Condition Equation

Set the molarity equation from step 2 equal to the molarity from step 3 to satisfy isotonic conditions: \[ \frac{1}{6} = \frac{50}{M_2} \]

Step 5: Solving for Molecular Mass of the Non-Volatile Solute

Cross-multiply and solve for \(M_2\): \[ M_2 = 50 \times 6 = 300 \, \text{g/mol} \]

Step 6: Choosing the Correct Answer

Comparing the calculated molecular mass (300 with the given options. The correct answer is option (d), 300 g/mol.

Key Concepts

Osmotic Pressure

In chemistry, osmotic pressure is a key concept that relates to how solutions behave in different environments. It's particularly important for understanding isotonic solutions, where two solutions have equal osmotic pressure.
Osmotic pressure is the pressure required to stop the flow of solvent across a semipermeable membrane. It depends primarily on the number of solute particles in the solution. Therefore, it is directly proportional to molarity (concentration of solute) and temperature, following the formula:

• \( \pi = i \cdot C \cdot R \cdot T \)

where \( \pi \) is the osmotic pressure, \( i \) is the van 't Hoff factor, \( C \) is the molarity, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
In the context of the original exercise, both solutions mentioned (the urea and non-volatile solute) maintain equal osmotic pressures when isotonic because they have the same molarity, given they are at the same temperature and pressure.

Molarity Calculation

Molarity is a vital unit of concentration in chemistry that tells you how much solute is present per unit volume of solution. It is measured in moles per liter (mol/L or M). It's calculated using the formula:

• \( M = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \times \frac{1}{\text{volume of solution (L)}} \)

To find the molarity of the urea solution, you divide the mass of urea (10 g) by its molar mass (60 g/mol) and adjust for the volume (1 liter). This yields a molarity of \( \frac{1}{6} \) mol/L.
For the non-volatile solute, we use the given 5% w/v concentration, converting it to a theoretical molarity to compare with the urea solution to achieve isotonic conditions.

van 't Hoff factor

The van 't Hoff factor, \( i \), is crucial in colligative properties and accounts for the number of particles a solute forms in solution. For non-electrolytes like urea and the non-volatile solute in the problem, which do not dissociate into ions,
\( i \) equals 1. This means each molecule of solute stays whole in solution, affecting the osmotic pressure linearly with the amount of solute added.
It's essential to distinguish between electrolytes and non-electrolytes here, since electrolytes would have \( i \) values greater than 1, reflecting the additional particles formed upon dissociation. In isotonic solutions, like the problem scenario, assuming \( i = 1 \) simplifies calculations when dealing with such non-electrolytic solutes.

Molecular Mass Determination

Determining molecular mass accurately is a common task in chemistry, especially in cases like the exercise where it involves solving concentration-related problems. You calculate molecular mass by comparing known concentrations of different solutions to ensure consistent osmotic pressures.
The molecular mass of the unknown non-volatile solute was deduced using isotonic conditions: setting molarity of the urea solution equal to that of the unknown solute. Through mathematical manipulation, you isolate and solve for the molecular mass:
- By equating the molarity formulas: \( \frac{1}{6} = \frac{50}{M_2} \).
- Solving gives \( M_2 = 300 \) g/mol.
Knowing molecular mass helps understand solute characteristics such as size and mass, crucial for predicting behavior in various chemical contexts.