The freezing point of water is 0.00°C at 1 atmosphere. A student dissolves 10.56 grams of aluminum acetate, Al(CH3COO)3 (204.1 g/mol), in 247.4 grams of water. Use the table of boiling and freezing point constants to answer the questions below. K, CC/m) Kf CC/m) Solvent Formula Water H20 0.512 1.86 Ethanol CH;CH,OH 1.22 1.99 Chloroform CHC1; 3.67 Benzene C6H6 2.53 5.12 Diethyl ether CH;CH,OCH,CH3 2.02 The molality of the solution is m. The freezing point of the solution is °C.

Biochemistry
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Chapter1: Biochemistry: An Evolving Science
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The freezing point of water is 0.00°C at 1 atmosphere.

A student dissolves 10.56 grams of aluminum acetate, Al(CH₃COO)₃ (204.1 g/mol), in 247.4 grams of water. Use the table of boiling and freezing point constants to answer the questions below.

| 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 molality of the solution is __________ m.

The freezing point of the solution is __________ °C.
Transcribed Image Text:The freezing point of water is 0.00°C at 1 atmosphere. A student dissolves 10.56 grams of aluminum acetate, Al(CH₃COO)₃ (204.1 g/mol), in 247.4 grams of water. Use the table of boiling and freezing point constants to answer the questions below. | 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 molality of the solution is __________ m. The freezing point of the solution is __________ °C.
Expert Solution
Step 1

molality = weight of solute*1000/gram molecular weight of solute* weight of solvent in g = 10.56*1000/(204.1*247.4) = 0.209m

 

 

Al(CH3COO)3 (aq) ----------> AI3+ (aq) + 3CH3C00-(aq)

i = 4

∆Tf = i*Kf*m

= 4*1.86*0.209= 1.55°C

∆Tf = Tf solvent - Tf solution

1.55= 0-Tf solution

Tf solution = -1.55°C

The freezing point of the solution = -1.55°C

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