**Boiling Point Elevation/Freezing Point Depression**\[\Delta T = m \cdot K\]where, for freezing point depression:\[\Delta T = T(\text{pure solvent}) - T(\text{solution})\]and for boiling point elevation:\[\Delta T = T(\text{solution}) - T(\text{pure solvent})\]- \(m\) = (# moles solute / Kg solvent)- \(K_b\) = boiling point elevation constant.- \(K_f\) = freezing point depression constant.\(K_b\) and \(K_f\) depend only on the SOLVENT. Below are some common values. Use these values for the calculations that follow.| Solvent | Formula | \(K_b\)(°C / m) | \(K_f\)(°C / m) ||--------------|---------------|------------------|------------------|| Water | H₂O | 0.512 | 1.86 || Ethanol | CH₃CH₂OH | 1.22 | 1.99 || Chloroform | CHCl₃ | 3.67 | || Benzene | C₆H₆ | 2.53 | 5.12 || Diethyl ether| CH₃CH₂OCH₂CH₃ | 2.02 | | The freezing point of water is 0.00 °C at 1 atmosphere.*Problem Statement:* How many grams of calcium nitrate (164.1 g/mol) must be dissolved in 218.0 grams of water to reduce the freezing point by 0.350 °C? Refer to the table for the necessary boiling or freezing point constant.*Data Table:* - **Solvent:** Different solvents and their properties- **Formula:** Chemical formula of each solvent- \( K_b \) \((\degree C/m)\): The boiling point elevation constant for each solvent- \( K_f \) \((\degree C/m)\): The freezing point depression constant for each solvent| Solvent | Formula | \( K_b \) \((\degree C/m)\) | \( K_f \) \((\degree C/m)\) ||----------------|--------------------|-----------------------------|-----------------------------|| Water | \( \text{H}_2\text{O} \) | 0.512 | 1.86 || Ethanol | \( \text{CH}_3\text{CH}_2\text{OH} \) | 1.22 | 1.99 || Chloroform | \( \text{CHCl}_3 \) | 3.67 | || Benzene | \( \text{C}_6\text{H}_6 \) | 2.53 | 5.12 || Diethyl ether | \( \text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3 \) | 2.02 | |The task is to calculate the mass of calcium nitrate needed using the properties of water from this table: \( K_f = 1.86 \degree C/m \).**Mass = \(\boxed{\phantom{g}}\)**

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Boiling Point Elevation/Freezing Point Depression AT = m K where, for freezing point depression: AT = T(pure solvent) - T(solution) and for boiling point elevation: AT T(solution) - T(pure solvent) m = (# moles solute / Kg solvent) Kb boiling point elevation constant. Kf = freezing point depression constant. Kb and Kf depend only on the SOLVENT. Below are some common values. Use these values for the calculations that follow. Solvent Kb(°C/m) Kf(°C / m) Water 0.512 1.86 Ethanol CH3CH₂OH 1.22 1.99 Chloroform CHCI 3 3.67 Benzene C6H6 2.53 Diethyl ether CH3CH₂OCH₂CH3 2.02 Formula H₂O 5.12

Boiling Point Elevation/Freezing Point Depression AT = m K where, for freezing point depression: AT = T(pure solvent) - T(solution) and for boiling point elevation: AT T(solution) - T(pure solvent) m = (# moles solute / Kg solvent) Kb boiling point elevation constant. Kf = freezing point depression constant. Kb and Kf depend only on the SOLVENT. Below are some common values. Use these values for the calculations that follow. Solvent Kb(°C/m) Kf(°C / m) Water 0.512 1.86 Ethanol CH3CH₂OH 1.22 1.99 Chloroform CHCI 3 3.67 Benzene C6H6 2.53 Diethyl ether CH3CH₂OCH₂CH3 2.02 Formula H₂O 5.12

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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