Problem 71 - A Complete Resource Book in Chemistry for JEE Main

Question

A solution is obtained by dissolving \(6 \mathrm{~g}\) of urea (mol. wt = 60) in a litre solution, another solution is prepared by dissolving \(34.2 \mathrm{~g}\) of cane sugar (mol. \(\mathrm{wt}=342\) ) in a litre of solution at the same temperature The lowering of vapour pressure in the first solution is (a) same as that of second solution (b) double that of second solution (c) half that of second solution (d) nearly one fifth of the second solution

Step-by-Step Solution

Answer

(a) The lowering of vapour pressure is the same for both solutions.

Step 1: Calculate Moles of Urea

Start by calculating the moles of urea. Given that the molecular weight of urea is 60, use the formula \( \text{moles} = \frac{\text{mass}}{\text{molecular weight}} \). Plug in the values for urea: \( \text{moles of urea} = \frac{6}{60} = 0.1 \) moles.

Step 2: Calculate Moles of Cane Sugar

Now, calculate the moles of cane sugar. The molecular weight of cane sugar is 342. Use the same formula: \( \text{moles of cane sugar} = \frac{34.2}{342} = 0.1 \) moles.

Step 3: Understand Vapour Pressure Lowering Concept

Lowering of vapour pressure is directly proportional to the number of moles of solute in a solution. Since both solutions have the same number of moles of solute (0.1 moles), the lowering of vapour pressure should be the same for both solutions.

Step 4: Evaluate the Answer

Since the number of moles of solute is the same in both solutions, the correct answer is that the lowering of vapour pressure in the first solution is the same as that of the second solution.

Key Concepts

Moles Calculation

To determine the number of moles in a solution, you need to understand how moles are calculated. The mole is a fundamental unit in chemistry that represents a specific quantity of any substance. It relates the mass of an element or compound to the number of atoms or molecules present. To calculate the moles of a substance, use the formula:

• Moles = \( \frac{\text{mass of the substance}}{\text{molecular weight}} \)

Let's break it down:- **Mass of the substance** is usually given in grams.- **Molecular weight** is the mass of one mole of a substance, usually given in grams per mole (g/mol). For instance, if you have 6 grams of urea with a molecular weight of 60 g/mol, you would have:\[\text{Moles of urea} = \frac{6}{60} = 0.1 \text{ moles}\]This formula is essential because it helps predict how substances will react in chemical reactions by knowing the number of moles.

Molecular Weight

Molecular weight quantifies the mass of a single molecule of a substance. It is often expressed in atomic mass units (amu) or grams per mole (g/mol). Calculating molecular weight is a crucial step in determining how much of a substance you need for a reaction. Molecular weight is calculated by summing the atomic weights of all atoms in a molecule. For example: - **Urea (NH2CONH2)** has a molecular weight of 60 g/mol. This is computed by adding the atomic weights of 4 hydrogen atoms, 2 nitrogen atoms, 1 carbon atom, and 1 oxygen atom. - For cane sugar (C12H22O11), each carbon, hydrogen, and oxygen atom contributes to a total molecular weight of 342 g/mol. Knowing molecular weight helps in:

• Calculating how many moles a given mass corresponds to using the moles formula.
• Determining the relationships between substances in reactions by comparing moles instead of masses directly.

Colligative Properties

Colligative properties are properties of solutions that depend on the number of solute particles but not on their identity. This concept is crucial in understanding vapor pressure lowering. Here are some key points about colligative properties:

• These include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.
• The change in a colligative property is directly proportional to the number of moles of solute added, independent of the type of solute.

For vapor pressure lowering, the important factor is the number of solute particles in the solution: - When a non-volatile solute is added to a solvent, it decreases the solvent's vapor pressure. - The more solute particles present, the greater the lowering of vapor pressure. In our exercise, both solutions had 0.1 moles of solute. Therefore, the vapor pressure lowering was equal, emphasizing that the molecular identity of the solvent—urea versus cane sugar—did not matter, only the number of solute particles did.