**Freezing Point Depression Experiment**The freezing point of water, \( \text{H}_2\text{O} \), is \( 0.000 \, ^\circ \text{C} \) at 1 atmosphere. The cryoscopic constant \( \text{K}_f(\text{water}) = 1.86 \, ^\circ \text{C/m} \).In a laboratory experiment, students synthesized a new compound and found that when \( 13.57 \) grams of the compound were dissolved in \( 272.6 \) grams of water, the solution began to freeze at \(-1.491 \, ^\circ \text{C}\). The compound was also found to be nonvolatile and a non-electrolyte.**Objective:**What is the molecular weight they determined for this compound?\[ \underline{\phantom{000}} \, \text{g/mol} \] The common laboratory solvent water is often used to purify substances dissolved in it. The vapor pressure of water is 23.76 mm Hg at 25 °C.In a laboratory experiment, students synthesized a new compound and found that when 153.6 grams of the compound were dissolved in 292.5 grams of water, the vapor pressure of the solution was 23.12 mm Hg. The compound was also found to be nonvolatile and a non-electrolyte.What is the molecular weight of this compound?(Water = H₂O = 18.02 g/mol)Molecular weight = [ ] g/mol

Answered: The freezing point of water, H₂O, is…

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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**Freezing Point Depression Experiment**

The freezing point of water, \( \text{H}_2\text{O} \), is \( 0.000 \, ^\circ \text{C} \) at 1 atmosphere. The cryoscopic constant \( \text{K}_f(\text{water}) = 1.86 \, ^\circ \text{C/m} \).

In a laboratory experiment, students synthesized a new compound and found that when \( 13.57 \) grams of the compound were dissolved in \( 272.6 \) grams of water, the solution began to freeze at \(-1.491 \, ^\circ \text{C}\). The compound was also found to be nonvolatile and a non-electrolyte.

**Objective:**

What is the molecular weight they determined for this compound?

\[ \underline{\phantom{000}} \, \text{g/mol} \]

The common laboratory solvent water is often used to purify substances dissolved in it. The vapor pressure of water is 23.76 mm Hg at 25 °C.

In a laboratory experiment, students synthesized a new compound and found that when 153.6 grams of the compound were dissolved in 292.5 grams of water, the vapor pressure of the solution was 23.12 mm Hg. The compound was also found to be nonvolatile and a non-electrolyte.

What is the molecular weight of this compound?

(Water = H₂O = 18.02 g/mol)

Molecular weight = [ ] g/mol

Expert Solution

Step 1: Colligative properties

Answer:

When a non-volatile is dissolved in a solvent it affects its chemical and physical properties. These properties depend on the number of solute particles dissolved in the solvent and are called as colligative properties. Lowering in vapor pression, depression in freezing point, elevation in boiling point etc. are the examples of colligative properties.

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