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

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**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             |                  |
Transcribed Image Text:**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 = \(\boxed{\phantom{g}}\)**
Transcribed Image Text: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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