Problem 60 - A Complete Resource Book in Chemistry for JEE Main

Question

The mass of glucose that should be dissolved in \(50 \mathrm{~g}\) of water in order to produce the same lowering of vapour pressure as is produced by dissolving \(1 \mathrm{~g}\) of urea in the same quantity of water is (a) \(1 \mathrm{~g}\) (b) \(3 \mathrm{~g}\) (c) \(6 \mathrm{~g}\) (d) \(18 \mathrm{~g}\)

Step-by-Step Solution

Answer

The correct mass of glucose required is 3 g.

Step 1: Understand the Given Problem

The problem asks us to find the mass of glucose needed to produce the same lowering of vapor pressure in 50 g of water as 1 g of urea would produce. We will compare these substances based on their molar mass because vapor pressure lowering is a colligative property, which depends on the number of solute particles.

Step 2: Calculate the Moles of Urea

First, we determine the number of moles of urea. The molar mass of urea ( ext{CH}_4 ext{N}_2 ext{O}) is approximately 60 g/mol. Therefore, the moles of urea is calculated as follows:\[ ext{Moles of urea} = \frac{1 ext{ g}}{60 ext{ g/mol}} = 0.0167 ext{ mol}\]

Step 3: Calculate Moles of Glucose Needed

To have the same colligative effect (lowering of vapor pressure), the solution must have the same number of moles of solute particles. Thus, we need 0.0167 moles of glucose. The molar mass of glucose ( ext{C}_6 ext{H}_{12} ext{O}_6) is approximately 180 g/mol.

Step 4: Find Mass of Glucose Needed

Now, we calculate the mass of glucose required:\[ ext{Mass of glucose} = ext{Moles of glucose} \times ext{Molar mass of glucose}\]\[ ext{Mass of glucose} = 0.0167 ext{ mol} \times 180 ext{ g/mol} = 3 ext{ g}\]

Step 5: Choose the Correct Answer

Based on our calculations, dissolving 3 g of glucose will create the same lowering of vapor pressure as dissolving 1 g of urea. Therefore, the correct answer is option (b) 3 g.

Key Concepts

Vapor pressure lowering

Vapor pressure lowering is a fascinating colligative property observed when a solute is added to a solvent, leading to a decrease in the solvent's vapor pressure. Colligative properties depend solely on the number of solute particles rather than their identity. When a non-volatile solute, like glucose or urea, is dissolved in a solvent such as water, it disrupts the ability of solvent molecules to escape into the vapor phase, thus lowering the vapor pressure of the solution.
In practical terms, this means that compared to pure water, the solution will have a lower tendency to evaporate. The amount by which the vapor pressure is lowered is directly proportional to the molality of the solute in the solution. Hence, to achieve the same vapor pressure lowering, the concentration of solute particles must be the same in both solutions.

Molarity

Molarity is a vital concept in chemistry, representing the concentration of a solute in a solution. It is expressed as the number of moles of solute per liter of solution. Symbolically, it is denoted as \(M\). Molarity provides a straightforward method to describe how concentrated a solution is.
In the context of the exercise, while we use the number of moles directly related to vapor pressure lowering, molarity offers insight into how these moles translate when dealing with larger volumes. If you were to prepare a solution with a known vapor pressure lowering, molarity would help determine how to scale up that solution accurately.

Molecular weight

Molecular weight, often synonymous with molar mass, is the sum of the atomic masses of all atoms in a molecule. It is usually measured in atomic mass units (amu) or grams per mole (g/mol). Knowing the molecular weight of substances, like glucose and urea in the exercise, is critical for converting between mass and moles. For example, in the provided exercise, we calculated the moles of urea using its molecular weight (60 g/mol). By doing this, we could compare how different solutes affect the colligative properties when dissolved in the same solvent. Molecular weight allows us to bridge the gap between the macroscopic mass measurements and the molecular world of chemical reactions.

Molar mass

Molar mass is a fundamental concept in stoichiometry, similar to molecular weight, but specifically refers to the mass of one mole of a substance, whether a molecule or an element. It is expressed in grams per mole (g/mol) and is calculated by summing the standard atomic masses of all components in a formula unit.
In the problem at hand, we used the molar mass of glucose (180 g/mol) to determine the mass required to equate the vapor pressure lowering to that of urea. By knowing the molar mass, we can convert moles of a substance into grams, facilitating the preparation of precise solutions for experiments and real-world usage.

Solution concentration

Solution concentration is an overarching concept that describes how much of a solute is present in a given quantity of solvent or solution. It can be expressed in various ways, such as molarity, molality, or mass percentage. This concentration affects the physical and chemical properties of the solution, such as boiling point, freezing point, and vapor pressure.
In our exercise, the focus was on achieving a specific concentration to induce a desired vapor pressure change. By understanding how to measure and manipulate solution concentration, chemists can predict and control the outcomes of experiments and industrial processes, ensuring that solutions are prepared with the right properties for their intended use.