/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 102 Mole fraction of solute in benze... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 102

Mole fraction of solute in benzene is \(0.2\), then the molality of the solution is a. 2 b. 4 c. \(3.2\) d. \(3.8\) 103\. Which one of the following aqueous solutions will exhibit highest boiling point? a. \(0.05 \mathrm{M}\) glucose b. \(0.01 \mathrm{M} \mathrm{KNO}_{3}\) c. \(0.015 \mathrm{M}\) urea d. \(0.01 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
Molality is 3.2. Highest boiling point: d (Na鈧係O鈧�).

Step by step solution

01

Understanding Mole Fraction

Given that the mole fraction of solute in benzene is 0.2, this means that 20% of the moles in the solution are the solute, and the remaining 80% are the solvent (benzene). We express this as a ratio of 0.2/0.8 for solute to solvent.
02

Calculate Number of Moles

Let the total number of moles be 1 mole for simplicity. According to the mole fraction, the number of moles of solute is 0.2, and the number of moles of benzene (solvent) is 0.8.
03

Calculate Molality

Molality ( extbf{m}) is defined as moles of solute per kilogram of solvent. Assuming the molar mass of benzene is approximately 78 g/mol, then: \[ \text{Mass of benzene} = 0.8 imes 78 = 62.4\text{ g} \] Convert this mass to kilograms: 62.4 g = 0.0624 kg. Therefore, the molality is: \[ \text{Molality (m)} = \frac{0.2}{0.0624} \approx 3.2 \]
04

Evaluate the Boiling Point Elevation

To find which solution exhibits the highest boiling point, we use the principle that boiling point elevation is proportionate to the van't Hoff factor (i) times molality (m).
05

Calculate Effect of Solutes

For the given solutions: (a) Glucose (non-electrolyte, i=1): 0.05 (b) KNO鈧� (i=2): 0.01 x 2 = 0.02 (c) Urea (non-electrolyte, i=1): 0.015 (d) Na鈧係O鈧� (i=3): 0.01 x 3 = 0.03, thus Na鈧係O鈧� has the highest van't Hoff factor and molality product.
06

Conclusion and Solution

Comparing the calculations, the solution with 0.01 M Na鈧係O鈧� (d) exhibits the highest boiling point due to its highest van't Hoff factor effect on boiling point elevation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mole Fraction
When talking about solutions in physical chemistry, the term 'Mole Fraction' is an essential concept. It tells us how much of one substance (solute) is present in a mixture compared to the total amount of all substances in the mixture. This is expressed as a number between 0 and 1, representing the ratio of moles of the solute to the total moles in the solution. If a solution has a mole fraction of solute equal to 0.2, this means 20% of the mixture is the solute, and the remaining 80% is composed of solvents like benzene.
  • Mole fraction is dimensionless, meaning it has no units.
  • It鈥檚 very useful for calculating other solution properties.
Knowing the mole fraction is crucial when determining other concentration measures, such as molality, and is often the starting point for calculating solution properties.
Molality
Molality is an important measure of concentration used in chemistry, especially when working with thermodynamic calculations. Unlike molarity, which depends on volume, molality (m) depends on the mass of the solvent, making it more reliable when temperature changes, as volume can expand or contract.To find the molality, divide the moles of solute by the kilograms of solvent. Let's say you have a mixture with 0.2 moles of solute and 0.0624 kg of benzene:
  • The formula is: \( \text{m} = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \).
  • So in this example, \( \text{m} = \frac{0.2}{0.0624} \approx 3.2 \text{ mol/kg} \).
Molality is especially useful in calculations that involve boiling point elevation and freezing point depression because these changes depend not on volume but on the mass of the solvent.
Boiling Point Elevation
Boiling point elevation occurs when a solute is added to a solvent, increasing the solution's boiling point above that of the pure solvent. This happens because solute particles disrupt the formation of vapor, making it harder for the solvent molecules to escape into the gas phase.
  • Boiling point elevation is quantified using the formula: \( \Delta T_b = i \times K_b \times m \), where \( \Delta T_b \) is the boiling point elevation, \( i \) is the van't Hoff factor, \( K_b \) is the ebullioscopic constant, and \( m \) is molality.
  • Every solute has a different effect on the boiling point, primarily based on how it dissociates in the solution, which is where the van't Hoff factor comes into play.
Understanding this concept is critical for predicting how different solutions will behave in processes that require precise temperature control, such as cooking or industrial chemical reactions.
Van't Hoff Factor
The Van't Hoff factor is a key element in physical chemistry, primarily used when studying colligative properties like boiling point elevation and freezing point depression. The van't Hoff factor (i) indicates how many particles a solute breaks into when dissolved.
  • For a non-electrolyte like glucose or urea, which doesn鈥檛 dissociate in solution, \( i = 1 \).
  • For electrolytes like KNO鈧� and Na鈧係O鈧�, the factor increases as each dissociates into multiple ions: \( i = 2 \) for KNO鈧� and \( i = 3 \) for Na鈧係O鈧�.
  • This factor is critical in accurately calculating changes in boiling and freezing points for solutions.
Considering the van't Hoff factor allows chemists to predict how solutes will behave when dissolved, influencing solution properties and making accurate predictions in experimental and applied chemistry scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(A): When \(\mathrm{CuSO}_{4} .5 \mathrm{H}_{2} \mathrm{O}\) is dissolved in water, the solution cools down and when \(\mathrm{CuSO}_{4}\) is dissolved in water, the solution gets heated up. (R): The dissolution of hydrated \(\mathrm{CuSO}_{4}\) is endothermic process because force of attraction decreases on dissolution whereas \(\mathrm{CuSO}_{4}\) (anhydrous) changes to hydrated and energy is released due to formation of new bonds.

A sugar syrup of weight \(214.2 \mathrm{~g}\) contains \(34.2 \mathrm{~g}\) of water. The molal concentration is a. \(0.55\) b. \(5.5\) c. 55 d. \(0.1\)

A solution is prepared by dissolving \(17.75 \mathrm{~g} \mathrm{H}_{2} \mathrm{SO}_{4}\) in enough water to make \(100.0 \mathrm{ml}\) of solution. If the density of the solution is \(1.1094 \mathrm{~g} / \mathrm{ml}\), what is the mole fraction \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the solution? a. \(1.338\) b. \(0.148\) c. \(3.38\) d. \(0.0338\)

Increasing the temperature of an aqueous solution will not cause a. decrease in molality b. decrease in molarity c. decrease in mole fraction d. decrease in \% (w/w).

When \(20 \mathrm{~g}\) of naphthoic acid \(\left(\mathrm{C}_{11} \mathrm{H}_{8} \mathrm{O}_{2}\right)\) is dissolved in \(50 \mathrm{~g}\) of benzene \(\left(\mathrm{Kf}=1.72 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\), a freezing point depression of \(2 \mathrm{~K}\) is observed. The Van't Hoff factor (i) is a. 1 b. \(0.5\) c. 3 d. 2 [IIT 2007] 209\. The Henry's law constant for the solubility of \(\mathrm{N}_{2}\) gas in water at \(298 \mathrm{~K}\) is \(1.0 \times 10^{5}\) atm. The mole fraction of \(\mathrm{N}_{2}\) in air is \(0.8\). The number of mole of \(\mathrm{N}_{2}\) from air dissolved in 10 moles of water at \(298 \mathrm{~K}\) and 5 atm pressure is a. \(4.0 \times 10^{-4}\) b. \(4.0 \times 10^{-5}\) c. \(5.0 \times 10^{-4}\) d. \(4.0 \times 10^{-6}\) [IIT 2009]
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.