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

Question

When \(18 \mathrm{~g}\) of glucose is dissolved in \(180 \mathrm{~g}\) of water then the mole fraction of glucose is (a) \(0.0099\) (b) \(0.0999\) (c) \(0.9999\) (d) \(0.9111\)

Step-by-Step Solution

Answer

(a) 0.0099

Step 1: Calculate the number of moles of glucose

To find the number of moles, use the formula \( ext{moles} = \frac{\text{mass}}{\text{molar mass}} \). The molar mass of glucose (\( \text{C}_6\text{H}_{12}\text{O}_6 \)) is approximately \( 180 \text{ g/mol} \). Thus, \( \text{moles of glucose} = \frac{18 \text{ g}}{180 \text{ g/mol}} = 0.1 \text{ mol} \).

Step 2: Calculate the number of moles of water

The molar mass of water (\( \text{H}_2\text{O} \)) is approximately \( 18 \text{ g/mol} \). Use the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Thus, \( \text{moles of water} = \frac{180 \text{ g}}{18 \text{ g/mol}} = 10 \text{ mol} \).

Step 3: Calculate the total moles in the solution

Add the number of moles of glucose and the number of moles of water: \( \text{Total moles} = 0.1 \text{ mol} + 10 \text{ mol} = 10.1 \text{ mol} \).

Step 4: Calculate the mole fraction of glucose

The mole fraction of glucose is calculated using \( \text{mole fraction} = \frac{\text{moles of glucose}}{\text{total moles in solution}} \). So, \( \text{Mole fraction of glucose} = \frac{0.1 \text{ mol}}{10.1 \text{ mol}} \approx 0.0099 \).

Step 5: Select the correct option

Based on our calculations, the correct answer is the option that matches our calculated mole fraction: \(0.0099\). Therefore, the correct answer is (a) \(0.0099\).

Key Concepts

Understanding Glucose Solutions

A glucose solution is a mixture of glucose (a simple sugar) and water. It is often used in various scientific, medical, and culinary applications. In a laboratory setting, knowing the concentration of glucose helps in predicting the behavior of the solution under different conditions. Concentrations can be expressed in various ways, such as molarity, molality, or mole fraction. In this context, we focus on mole fraction, which is a way of expressing the concentration as a ratio of moles of solute to total moles of the solution. This gives insights into the proportion of glucose compared to the entire solution.

Molar Mass Calculation

Calculating the molar mass is crucial for converting between mass and moles, which are essential for many chemical calculations. The molar mass of a substance is the mass of one mole of its molecules. For glucose, with a chemical formula of \(C_6H_{12}O_6\), you need to consider the atomic masses of carbon, hydrogen, and oxygen.

• Carbon \( (C) \) has an atomic mass of about 12 g/mol.
• Hydrogen \( (H) \) has an atomic mass of about 1 g/mol.
• Oxygen \( (O) \) has an atomic mass of about 16 g/mol.

To find the molar mass of glucose:
6 carbon atoms contribute \(6 \times 12 = 72\) g/mol,
12 hydrogen atoms contribute \(12 \times 1 = 12\) g/mol,
and 6 oxygen atoms contribute \(6 \times 16 = 96\) g/mol.
Adding these up results in the molar mass of glucose being \(72 + 12 + 96 = 180\) g/mol. This value is then used to calculate moles from a given mass of glucose.

Mole Calculation Techniques

Understanding moles is pivotal in chemistry. A mole is a fundamental concept used to quantify amounts of a substance based on its constituent particles, such as molecules, atoms, or ions. The formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \) allows us to calculate the number of moles from a given mass.

In performing the exercise,the mass of glucose given is 18 g. Using its molar mass of 180 g/mol, the moles of glucose are calculated as:\[\text{moles of glucose} = \frac{18 \text{ g}}{180 \text{ g/mol}} = 0.1 \text{ mol}.\]
Likewise, for water, the mass is 180 g, and with a molar mass of 18 g/mol, we find:\[\text{moles of water} = \frac{180 \text{ g}}{18 \text{ g/mol}} = 10 \text{ mol}.\]
This methodical approach allows us to understand the amount of each component in a solution.

Exploring Solubility of Glucose

Solubility indicates how much of a solute can dissolve in a solvent at a given temperature and pressure. In this exercise, the focus is primarily on understanding how concentrations like mole fraction relate to solubility.

Glucose is considered highly soluble in water because it can dissolve in large amounts. This property is due to its polar nature and the high polarity of water, allowing them to interact and mix well. Understanding solubility is crucial because it affects how solutions are prepared and used in various applications.

• For example, in medical and dietary contexts, glucose solutions are used intravenously, requiring precise concentration measurements to ensure effectiveness and safety.
• In laboratory settings, knowing solubility helps in preparing solutions with desired concentrations.

These insights into the concept of solubility broaden our understanding of how solutions behave and their practical applications.