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 nonvolatile solute is 300 g/mol (option d).

Step 1: Understand the isotonic condition

Two solutions are isotonic when they have the same osmotic pressure. The osmotic pressure (ID) of a solution can be calculated using the formula =RT, where  is the molarity (/molar volume),  is the gas constant, and  is the temperature in Kelvin.

Step 2: Calculate molarity of urea solution

For the urea solution, the mass is 10 g, and the molecular mass is 60 g/mol. Thus, the number of moles of urea is \( n = \frac{10}{60} \) mol. In 1 , the molarity (\( C_1 \)) is \( \frac{10}{60 \times 1} = \frac{1}{6} \) mol/dm³.

Step 3: Set molarity equation for unknown solute

For the nonvolatile solute having a 5% solution, it means 5 g of solute is dissolved in 100 mL (0.1 dm³) of solution. Let the molecular mass of the solute be \( M_x \) g/mol. The number of moles is \( \frac{5}{M_x} \). Therefore, the molarity (\( C_2 \)) is \( \frac{5}{M_x \times 0.1} = \frac{50}{M_x} \) mol/dm³.

Step 4: Equate Molarities

In isotonic solutions, molarity is the same, so set \( C_1 = C_2 \). Therefore, \( \frac{1}{6} = \frac{50}{M_x} \).

Step 5: Solve for molecular mass

Cross-multiply to solve for \( M_x \): \( M_x = 50 \times 6 \). Thus, \( M_x = 300 \) g/mol.

Key Concepts

Isotonic Solutions

When two solutions have the same osmotic pressure, they are called isotonic. This simply means that they exert the same pressure when separated by a semipermeable membrane. Osmotic pressure is a crucial concept in chemistry and is vital in understanding how solutions behave in different environments.
Osmotic pressure can be thought of as the force needed to stop the diffusion of water across the membrane. It depends on the concentration of solute particles in a solution and not on the type of particles. The formula for calculating osmotic pressure is given by \[ \Pi = nRT/V = CRT \]where:

• \(\Pi\) is the osmotic pressure
• \(n\) is the number of moles of solute
• \(R\) is the gas constant
• \(T\) is the temperature in Kelvin
• \(C\) is the molarity of the solution
• \(V\) is the volume of the solution

Understanding isotonic solutions is useful in many fields, especially in medicine, where it's important to ensure solutions around cells do not cause them to shrink or burst.

Molarity Calculation

Molarity is a way to express how concentrated a solute is within a solution. It is defined by the number of moles of solute per liter (dm³) of solution. Calculating molarity is an essential skill in chemistry as it helps in determining the concentration of solutions.
For example, if you have 10 grams of urea in one liter of solution with a molecular mass of urea as 60 g/mol, then the number of moles of urea is calculated as \[ n = \frac{10}{60} \] mol. Consequently, the molarity of the urea solution, \(C_1\), becomes:\[ C_1 = \frac{1}{6} \] mol/dm³.
This process of calculating molarity allows chemists to precisely adjust solution concentrations for a variety of reactions and experiments.
Remember, if the volume of the solution changes, so does the molarity, because molarity relies on the volume of the solution.

Molecular Mass Determination

Determining the molecular mass of an unknown solute is often a necessary step in identifying chemical compounds. Molecular mass can be deduced from various experimental techniques, one of which is through osmotic pressure in isotonic solutions.
In our exercise, after obtaining the molarity of the urea solution, we use the isotonic property to find the molarity of the unknown solution. This allows us to set up an equation to solve for the molecular mass of the unknown solute, \(M_x\).
For a 5% solution of the unknown, \(C_2\) is calculated as:\[ C_2 = \frac{50}{M_x} \] mol/dm³,where the assumption is that 5 grams of solute are dissolved in 100 mL of solution.
By equating \(C_1\) and \(C_2\) because they are isotonic, we cross-multiply to solve for \(M_x\). The result is\[ M_x = 50 \times 6 = 300 \]g/mol, giving us the molecular mass of the unknown solute. This method is both straightforward and practical when dealing with isotonic solutions.