Explanation Given: Freezing point of 0.1000 M solution of H 2 So 4 = 272.76 K Molality of H 2 So 4 = 0.1000 M Formula used: ∆ T f = i×m×K f ------(1) Calculation: Freezing point depression = (273.15 − 272.76)° C = 0.39° C Since, Molal depression in freezing point constsnt = 1.82° C/m Upon substituting the data into the equation (1) 0.39°C=i×0.1000 m×1.82∆ C/m Now vant hoff factor may be calculated as i= 0 .39 o C 0 .1000 m×1 .82 o C/m i = 2.1428 For, Dissociation of H 2 SO 4 (1 mol), we will see the table as below H 2 SO 4 (aq)⇌H + +H So 4 2 − 0.1000 0 0 -x x x 0...

Pearson eText for Chemistry: structures and Properties -- Instant Access (Pearson+)

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ISBN: 9780136951537

Author: Nivaldo Tro

Publisher: PEARSON+

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Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to a…

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Chapter 13, Problem 127E

Interpretation Introduction

Interpretation:

“ molality of ${So}_{4}^{2 -}$ ” should be identified.

Concept introduction:

molality of a solution is the number of moles divided by mass of solvent in kilogram and can be expressed as follows;

$m = \frac{n}{mass of solvent (kg)}$

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Chapter 13, Problem 126E

Chapter 13, Problem 128E

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Chapter 13 Solutions

Pearson eText for Chemistry: structures and Properties -- Instant Access (Pearson+)

Ch. 13 - What is a solution? What are the solute and...Ch. 13 - What does it mean when we say that a substance is...Ch. 13 - Why do two ideal gases thoroughly mix when...Ch. 13 - Prob. 4E Ch. 13 - Prob. 5E Ch. 13 - Explain how the relative strengths of...Ch. 13 - What does the statement like dissolves like mean...Ch. 13 - Prob. 8E Ch. 13 - What is the heat of hydration(Hhydration)? How...Ch. 13 - Prob. 10E

Ch. 13 - How does temperature affect the solubility of a...Ch. 13 - Prob. 12E Ch. 13 - How does pressure affect the solubility of a gas...Ch. 13 - What is Henry’s law? For what kinds of...Ch. 13 - Prob. 15E Ch. 13 - How are parts by mass and parts by volume used in...Ch. 13 - Prob. 17E Ch. 13 - What is Raoult’s law? For what kind of...Ch. 13 - Explain the difference between an ideal and a...Ch. 13 - Prob. 20E Ch. 13 - Prob. 21E Ch. 13 - What are colligative properties?Ch. 13 - Prob. 23E Ch. 13 - Explain the significance of the van’t Hoff factor...Ch. 13 - Prob. 25E Ch. 13 - Pick an appropriate solvent from Table 13.3 to...Ch. 13 - Which molecule would you expect to be more soluble...Ch. 13 - Prob. 28E Ch. 13 - Prob. 29E Ch. 13 - Prob. 30E Ch. 13 - When ammonium chloride (NH4Cl) is dissolved in...Ch. 13 - Prob. 32E Ch. 13 - Prob. 33E Ch. 13 - Use the given data to calculate the heats of...Ch. 13 - Lithium iodide has a lattice energy of...Ch. 13 - Prob. 36E Ch. 13 - A solution contains 25 g of NaCl per 100.0 g of...Ch. 13 - A solution contains 32 g of KNO3 per 100.0 g of...Ch. 13 - Prob. 39E Ch. 13 - A KCI solution containing 42 g of KCI per 100.0 g...Ch. 13 - Some laboratory procedures involving...Ch. 13 - A person preparing a fish tank fills the tank with...Ch. 13 - Prob. 43E Ch. 13 - Scuba divers breathing air at increased pressure...Ch. 13 - Calculate the mass of nitrogen dissolved at room...Ch. 13 - Use Henry’s law to determine the molar solubility...Ch. 13 - An aqueous NaCl solution is made using 112 g of...Ch. 13 - Prob. 48E Ch. 13 - To what volume should you dilute 50.0 mL of a...Ch. 13 - Prob. 50E Ch. 13 - Silver nitrate solutions are used to plate silver...Ch. 13 - Prob. 52E Ch. 13 - Prob. 53E Ch. 13 - Prob. 54E Ch. 13 - You can purchase nitric acid in a concentrated...Ch. 13 - You can purchase hydrochloric acid in a...Ch. 13 - Prob. 57E Ch. 13 - Prob. 58E Ch. 13 - Prob. 59E Ch. 13 - Prob. 60E Ch. 13 - Prob. 61E Ch. 13 - Prob. 62E Ch. 13 - Prob. 63E Ch. 13 - Prob. 64E Ch. 13 - A beaker contains 100.0 mL of pure water. 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“ molality of So 4 2 − ” should be identified. Concept introduction: molality of a solution is the number of moles divided by mass of solvent in kilogram and can be expressed as follows; m = n mass of solvent (kg)

Solution Summary: The author explains that molality of a solution is the number of moles divided by mass of solvent in kilogram.

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Interpretation:

“ molality of ${So}_{4}^{2 -}$ ” should be identified.

Concept introduction:

molality of a solution is the number of moles divided by mass of solvent in kilogram and can be expressed as follows;

$m = \frac{n}{mass of solvent (kg)}$

Want to see the full answer?

Chapter 13 Solutions

“ molality of So 4 2 − ” should be identified. Concept introduction: molality of a solution is the number of moles divided by mass of solvent in kilogram and can be expressed as follows; m = n mass of solvent (kg)

Solution Summary: The author explains that molality of a solution is the number of moles divided by mass of solvent in kilogram.

Concept explainers

Videos

Interpretation: “ molality of So42− ” should be identified. Concept introduction: molality of a solution is the number of moles divided by mass of solvent in kilogram and can be expressed as follows; m = nmass of solvent (kg)

Want to see the full answer?

Chapter 13 Solutions

Interpretation:

“ molality of ${So}_{4}^{2 -}$ ” should be identified.

Concept introduction:

molality of a solution is the number of moles divided by mass of solvent in kilogram and can be expressed as follows;

$m = \frac{n}{mass of solvent (kg)}$