Problem 161 - A Complete Resource Book in Chemistry for JEE Main

Question

A \(5.25 \%\) solution of a substance is isotonic with a \(1.5 \%\) solution of urea (molar mass \(=60 \mathrm{~g} \mathrm{~mol}^{-1}\) ) in the same solvent. If the densities of both the solutions are assumed to be equal to \(1.0 \mathrm{~g} \mathrm{~cm}^{-3}\), molar mass of the substance will be (a) \(115.0 \mathrm{~g} \mathrm{~mol}^{-1}\) (b) \(105.0 \mathrm{~g} \mathrm{~mol}^{-1}\) (c) \(210.0 \mathrm{~g} \mathrm{~mol}^{-1}\) (d) \(90.0 \mathrm{~g} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The molar mass of the substance is \(210.0 \ \text{g/mol}\).

Step 1: Understanding Isotonic Solutions

Two solutions are isotonic when they have the same osmotic pressure. In this problem, we have a solution of a substance and a solution of urea, which are isotonic. Therefore, the osmotic pressures \( \Pi \) of both solutions can be set equal: \( \Pi_1 = \Pi_2 \).

Step 2: Using the Formula for Osmotic Pressure

The formula for osmotic pressure is given by \( \Pi = iCRT \), where \( i \) is the van't Hoff factor (assumed to be 1 for non-electrolyte solutions), \( C \) is the molarity, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. Since \( R \) and \( T \) are identical for both solutions and \( i=1 \), the equation simplifies to \( C_1 = C_2 \) (under equal conditions of temperature and pressure).

Step 3: Determine Molarity of the Urea Solution

For a \(1.5\%\) solution of urea, there are \(1.5\) grams of urea in \(100\) mL of the solution. Since density is \(1.0\ \text{g/cm}^3\), \(100\) mL has a mass of \(100\) grams. The moles of urea can be calculated as \(\frac{1.5}{60.0} = 0.025 \text{ moles}\). Thus, molarity \( C \) of the urea solution is \(0.25\ \text{mol/L}\).

Step 4: Using the Molarity to Find Molar Mass of the Substance

For a \(5.25\%\) solution, there are \(5.25\) grams in \(100\) mL, which is \(0.0525\ \text{grams/mL}\). Given the density, the same volume applies. The molarity of the substance solution is also \(0.25\ \text{mol/L}\) due to isotonic conditions. The molar mass \( M \) can be calculated as \(\frac{5.25}{0.25} = 21.0 \ \text{g/mol}\).

Step 5: Solving for Molar Mass of the Substance

There seems to be inconsistency in our calculation; let’s recompute: For molarity to equal \(0.25\ \text{mol/L}\):\[\text{Molar mass of the substance}= \frac{\text{mass of the substance} (\text{g})}{\text{moles of the substance}}\Rightarrow\\text{Molar mass} = \frac{5.25}{0.025} = 210.0 \ \text{g/mol}\]

Key Concepts

Understanding Isotonic Solutions

Isotonic solutions are an important concept in chemistry, particularly in understanding osmotic pressure. Two solutions are considered isotonic when they have the same osmotic pressure, which is why they don't exchange water when separated by a semi-permeable membrane. In the context of this exercise, the isotonic nature is established between a solution with a 5.25% concentration of an unknown substance and a 1.5% solution of urea. Both are in the same solvent and have a density of 1.0 g/cm³. This means both solutions exert the same osmotic pressure. Understanding isotonic conditions helps us relate the molarity of the two solutions under identical temperature and pressure conditions.

Delving into Molarity

Molarity is one of the ways to express the concentration of a solution. It defines how many moles of solute are present in one liter of solution. In this exercise, the molarity of the urea solution is calculated first. For the 1.5% urea solution, this means there are 1.5 grams of urea in 100 mL of solution. Considering the density is 1.0 g/cm³, we calculate the molarity as follows:

• You determine the moles of urea with its molar mass: \( \frac{1.5}{60.0} = 0.025 \) moles.
• The molarity, \( C \), becomes 0.25 mol/L because 0.025 moles are present in 0.1 liters.

By setting the urea solution's molarity equal to the unknown substance's solution, they are said to be isotonic, indicating their equivalence in osmotic pressure under the same conditions.

Calculating Molar Mass Accurately

Molar mass calculation is essential when dealing with solutions and assessing their properties. In this exercise, we want to find the molar mass of the unknown substance that's isotonic to a urea solution. Knowing the molarity can resemble us the steps required. For the 5.25% solution of the unknown substance:

• There are 5.25 grams of the substance in 100 mL (0.1 L), equating to \( 0.0525 \) g/mL given the 1.0 g/cm³ density.
• By finding the molarity as 0.25 mol/L, due to isotonic conditions, we can use the formula: \( \text{Molar mass} = \frac{\text{grams of substance}}{\text{moles}} \).
• The molar mass is found by \( \frac{5.25}{0.025} = 210.0 \) g/mol, showcasing the relation of mass to mol and aligning with isotonic principles.

Exploring the Properties of Solutions

The properties of solutions are fundamental to many chemical calculations and experiments. They include properties like concentration, density, and osmosis. In this exercise, understanding how isotonic solutions relate leverages osmotic pressure properties and concentration measurements, including:

• Concentration: It reflects how much solute is placed in a solvent, significant for calculating molarity and understanding osmotic equilibriums.
• Osmotic Pressure: This defines the tendency of solvent molecules to move through a semi-permeable membrane, seeking equal concentrations on both sides.

Each property deeply interconnects with the other, helping comprehend how substances in solution behave when they are isotonic with another solution. Mastery of these basics helps deconstruct complex problems, like identifying the molar mass of an unknown substance as in this exercise.