/*! 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 39 The molar mass of substance whic... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 39

The molar mass of substance which forms \(7.0 \%\) (by mass) solution in water which freezes at \(-0.93^{\circ} \mathrm{C}\). The cryoscopic constant of water is \(1.86^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}\). (a) \(140 \mathrm{~g} \mathrm{~mol}^{-1}\) (b) \(150.5 \mathrm{~g} \mathrm{~mol}^{-1}\) (c) \(160 \mathrm{~g} \mathrm{~mol}^{-1}\) (d) \(155 \mathrm{~g} \mathrm{~mol}^{-1}\)

Short Answer

Expert verified
The molar mass of the solute is approximately 150.5 g/mol.

Step by step solution

01

- Understanding the Problem

We need to determine the molar mass of a solute that is present at a concentration of 7.0% by mass in a solution. The solution freezes at -0.93 degrees Celsius, which is lower than the freezing point of pure water (0 degrees Celsius). This depression in the freezing point is related to the molality of the solution and the cryoscopic constant for water.
02

- Calculate Molality of the Solution

Using the freezing point depression formula \(\Delta T_f = i \cdot K_f \cdot m\), where \(\Delta T_f\) is the freezing point depression, \(i\) is the van't Hoff factor (equal to 1 for non-electrolytes), \(K_f\) is the cryoscopic constant, and \(m\) is the molality of the solution, we can solve for the molality since \(\Delta T_f\), \(i\), and \(K_f\) are known. The depression of freezing point is given as the difference between the freezing points of pure solvent and the solution, which is \(\Delta T_f = 0 - (-0.93^\circ C) = 0.93^\circ C\).
03

- Calculate the Molality (m)

Solving for molality (m) with the cryoscopic constant of water (\(K_f = 1.86^\circ C \cdot kg/mol\)) and the given \(\Delta T_f = 0.93^\circ C\), the formula becomes \(0.93 = 1 \cdot 1.86 \cdot m\). So, \(m = \frac{0.93}{1.86}\).
04

- Calculate the Molar Mass of the Solute

Molality (m) is defined as the number of moles of solute per kilogram of solvent. Since 100 g of the solution contains 7 g of solute and 93 g of solvent, the mass of the solvent in kilograms is 0.093 kg. Using the molality obtained from Step 3, we can calculate the moles of solute, and from there, the molar mass of the solute.
05

- Find Moles of Solute

Using the calculated molality (m), the number of moles of solute in 0.093 kg of solvent is the product of molality and mass of solvent in kilograms, \( moles = molality \times mass \).
06

- Calculate Molar Mass

Finally, to get the molar mass, we divide the mass of solute (in grams) by the moles of solute we've just found.

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.

Freezing Point Depression
Freezing point depression occurs when a solute is added to a solvent, resulting in the lowering of the freezing temperature of the solution compared to the pure solvent. This is a colligative property, which means it depends on the number of solute particles in the solution and not on their identity. The extent of the freezing point depression can be calculated using the formula ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline \(\Delta T_f = i \cdot K_f \cdot m\), where \(\Delta T_f\) is the change in freezing point, \(i\) is the van鈥檛 Hoff factor indicating the number of particles the solute forms in the solution, \(K_f\) is the cryoscopic constant, which is specific to the solvent, and \(m\) is the molality of the solution.
Molality
Molality is a concentration unit that expresses the amount of solute in moles per kilogram of solvent, symbolized as m. It's different from molarity, which is moles per liter of solution. To calculate molality, use the formula ewline ewline ewline ewline \(m = \frac{{\text{{moles of solute}}}}{{\text{{mass of solvent in kilograms}}}}\). ewline ewline ewline ewline To find the molality in our exercise, we take the mass of the solvent (water in this case) in kilograms and the moles of solute, which we can derive from the given percentage composition and total mass of the solution.
Molar Mass Calculation
The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). It can be calculated by dividing the mass of the solute by the number of moles of the solute. The formula is ewline ewline ewline ewline \(\text{{Molar Mass}} = \frac{{\text{{mass of solute (g)}}}}{{\text{{moles of solute}}}}\). ewline ewline ewline ewline When a solution's composition is known, as in our exercise, the molar mass can be found by determining the moles of solute present in a known mass of solvent and then using the mass of solute present in the entire solution.
van't Hoff Factor
The van鈥檛 Hoff factor, denoted as \(i\), is a measure of the effect of solute particles on the colligative properties of a solution. For nonelectrolytes, substances that do not dissociate into ions in solution, the van鈥檛 Hoff factor is 1. For electrolytes, which dissociate into ions, the value of \(i\) is typically greater than 1, based on the number of particles the compound dissociates into. ewline ewline ewline ewline In freezing point depression and other colligative property calculations, the van鈥檛 Hoff factor is used in the formula to adjust for the number of particles the solute contributes to the solution and thus its impact on the solution's properties.

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

The degree of dissociation \((\alpha)\) of a weak electrolyte, \(A_{x} B_{y}\), is related to Van't Hoff factor (i) by the expression (a) \(\alpha=\frac{i-1}{x+y-1}\) (b) \(\alpha=\frac{i-1}{x+y+1}\) (c) \(\alpha=\frac{x+y-1}{i-1}\) (d) \(\alpha=\frac{x+y+1}{i-1}\)

\(\mathrm{PtCl}_{4} 6 \mathrm{H}_{2} \mathrm{O}\) can exist as a hydrated complex. \(1.0\) molal aqueous solution has depression in freezing point of \(3.72^{\circ} \mathrm{C}\). Assume \(100 \%\) ionization and \(K_{\ell}\) of water \(=1.86^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg}\). The complex is (a) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)\right] \mathrm{Cl}_{4}\) (b) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (c) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \mathrm{Cl} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) (d) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{4}\right] \cdot 4 \mathrm{H}_{2} \mathrm{O}\)

A liquid mixture of ' \(\mathrm{A}\) ' and 'B' (assume ideal solution) is placed in a cylinder and piston arrangement. The piston is slowly pulled out isothermally so that the volume of liquid decreases and that of the vapour increases. At the instant when the quantity of the liquid still remaining is negligibly small, the mole fraction of 'A' in the vapour is \(0.4\). If \(P_{\mathrm{A}}^{\circ}=0.4 \mathrm{~atm}\), \(P_{\mathrm{B}}^{\circ}=1.2 \mathrm{~atm}\) at this temperature, the total pressure at which the liquid has almost evaporated, is (a) \(0.667\) atm (b) \(1.5 \mathrm{~atm}\) (c) \(0.8 \mathrm{~atm}\) (d) \(0.545 \mathrm{~atm}\)

The boiling point and freezing point of a solvent 'A' are \(90.0^{\circ} \mathrm{C}\) and \(3.5^{\circ} \mathrm{C}\), respectively. \(K_{\mathrm{f}}\) and \(K_{\mathrm{b}}\) values of the solvent are \(17.5\) and \(5.0 \mathrm{~K}-\mathrm{kg} / \mathrm{mol}\), respectively. What is the boiling point of a solution of 'B' (non-volatile, nonelectrolyte solute) in 'A', if the solution freezes at \(2.8^{\circ} \mathrm{C} ?\) (a) \(90.0^{\circ} \mathrm{C}\) (b) \(89.8^{\circ} \mathrm{C}\) (c) \(90.2^{\circ} \mathrm{C}\) (d) \(90.7^{\circ} \mathrm{C}\)

A liquid is in equilibrium with its vapour at its boiling point. On average, the molecules in the two phases have equal (a) potential energy (b) intermolecular force (c) kinetic energy (d) total energy
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

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.