The freezing point of ethanol, CH3CH₂OH, is -117.300 °C at 1 atmosphere. Kf(ethanol) = 1.99 °C/m In a laboratory experiment, students synthesized a new compound and found that when 13.92 grams of the compound were dissolved in 224.6 grams of ethanol, the solution began to freeze at -117.438 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound? g/mol

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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**Freezing Point Depression Experiment**

The freezing point of ethanol, \( \text{CH}_3\text{CH}_2\text{OH} \), is \(-117.300 \, ^\circ\text{C}\) at 1 atmosphere. The cryoscopic constant (\( K_f \)) for ethanol is \( 1.99 \, ^\circ\text{C/m} \).

In a laboratory experiment, students synthesized a new compound and observed that when \( 13.92 \) grams of the compound were dissolved in \( 224.6 \) grams of ethanol, the solution began to freeze at \(-117.438 \, ^\circ\text{C}\). The compound was also found to be nonvolatile and a non-electrolyte.

**Question:**
What is the molecular weight they determined for this compound?

\[ \_\_\_\_\_ \, \text{g/mol} \]
Transcribed Image Text:**Freezing Point Depression Experiment** The freezing point of ethanol, \( \text{CH}_3\text{CH}_2\text{OH} \), is \(-117.300 \, ^\circ\text{C}\) at 1 atmosphere. The cryoscopic constant (\( K_f \)) for ethanol is \( 1.99 \, ^\circ\text{C/m} \). In a laboratory experiment, students synthesized a new compound and observed that when \( 13.92 \) grams of the compound were dissolved in \( 224.6 \) grams of ethanol, the solution began to freeze at \(-117.438 \, ^\circ\text{C}\). The compound was also found to be nonvolatile and a non-electrolyte. **Question:** What is the molecular weight they determined for this compound? \[ \_\_\_\_\_ \, \text{g/mol} \]
**Boiling Point Elevation/Freezing Point Depression**

\[ \Delta T = m \, K \]

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        | \( \text{H}_2\text{O} \)      | 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            |                 |
Transcribed Image Text:**Boiling Point Elevation/Freezing Point Depression** \[ \Delta T = m \, K \] 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 | \( \text{H}_2\text{O} \) | 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 | |
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