**Cryoscopic Constant and Freezing Point Depression***The freezing point of water is 0.00°C at 1 atmosphere.***Problem Statement:**How many grams of ammonium chloride (53.50 g/mol) must be dissolved in 211.0 grams of water to reduce the freezing point by 0.500°C? Refer to the table for the necessary boiling or freezing point constant.**Solvent Properties:**| Solvent | Formula | \( K_b \) (°C/m) | \( K_f \) (°C/m) ||-----------------|--------------------|------------------|------------------|| Water | H\(_2\)O | 0.512 | 1.86 || Ethanol | CH\(_3\)CH\(_2\)OH | 1.22 | 1.99 || Chloroform | CHCl\(_3\) | 3.67 | - || Benzene | C\(_6\)H\(_6\) | 2.53 | 5.12 || Diethyl ether | CH\(_3\)CH\(_2\)OCH\(_2\)CH\(_3\) | 2.02 | - |___**Calculation to be done:**\[ \_\_\_\_\_\_\_\, \text{g ammonium chloride} \]**Explanation:**To solve this problem, we will use the concept of freezing point depression. According to the formula:\[ \Delta T_f = K_f \cdot m \]where:- \( \Delta T_f \) is the decrease in freezing point (0.500°C),- \( K_f \) is the cryoscopic constant of water (1.86°C kg/mol),- \( m \) is the molality of the solution.Next, we find the molality (\(m\)) using:\[ m = \frac{\Delta T_f}{K_f} \]Then molality is converted to grams of solute using:\[ \text{grams of solute} = m \cdot \text{molar mass of solute} \cdot \text{kg of solvent} \]**Steps Involved:**1. Calculate the molality \(m\).2. Convert

Given Mass of water = 211 gram = 0.211 Kg Molar mass of Ammonium…

The freezing point of water is 0.00°C at 1 atmosphere. How many grams of ammonium chloride (53.50 g/mol), must be dissolved in 211.0 grams of water to reduce the freezing point by 0.500°C ? Refer to the table for the necessary boiling or freezing point constant. Solvent Formula K, (°C/m) K¢(°C/m) Water H2O 0.512 1.86 Ethanol CH;CH,OH 1.22 1.99 Chloroform CHCI3 3.67 Benzene 2.53 5.12 Diethyl ether CH3CH,OCH,CH3 2.02 g ammonium chloride.

The freezing point of water is 0.00°C at 1 atmosphere. How many grams of ammonium chloride (53.50 g/mol), must be dissolved in 211.0 grams of water to reduce the freezing point by 0.500°C ? Refer to the table for the necessary boiling or freezing point constant. Solvent Formula K, (°C/m) K¢(°C/m) Water H2O 0.512 1.86 Ethanol CH;CH,OH 1.22 1.99 Chloroform CHCI3 3.67 Benzene 2.53 5.12 Diethyl ether CH3CH,OCH,CH3 2.02 g ammonium chloride.

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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