5 mL of 0.3 M acetic acid is shaken with 5 g of activated charcoal. The concentration of acetic acid is reduced to 1/3 of the original molarity. The weight of acetic acid adsorbed per gram of charcoal is:

Answer

To solve the problem step by step, we will follow the calculations as outlined in the video transcript.

### Step 1: Calculate the initial millimoles of acetic acid.
We know the volume of acetic acid solution and its molarity. The formula to calculate millimoles is:

\[ \text{Millimoles} = \text{Volume (mL)} \times \text{Molarity (M)} \]

Given:
- Volume = 5 mL
- Molarity = 0.3 M

Calculating:

\[ \text{Millimoles} = 5 \, \text{mL} \times 0.3 \, \text{M} = 1.5 \, \text{mmol} \]

### Step 2: Calculate the final millimoles of acetic acid after adsorption.
According to the problem, the concentration of acetic acid is reduced to \( \frac{1}{3} \) of the original molarity.

Calculating the final concentration:

\[ \text{Final concentration} = \frac{1}{3} \times 0.3 \, \text{M} = 0.1 \, \text{M} \]

Now, we calculate the final millimoles in 5 mL:

\[ \text{Final millimoles} = 5 \, \text{mL} \times 0.1 \, \text{M} = 0.5 \, \text{mmol} \]

### Step 3: Calculate the millimoles of acetic acid adsorbed.
To find the amount of acetic acid adsorbed, we subtract the final millimoles from the initial millimoles:

\[ \text{Millimoles adsorbed} = \text{Initial millimoles} - \text{Final millimoles} \]

Calculating:

\[ \text{Millimoles adsorbed} = 1.5 \, \text{mmol} - 0.5 \, \text{mmol} = 1.0 \, \text{mmol} \]

### Step 4: Convert millimoles to moles.
To convert millimoles to moles, we divide by 1000:

\[ \text{Moles adsorbed} = \frac{1.0 \, \text{mmol}}{1000} = 1.0 \times 10^{-3} \, \text{mol} \]

### Step 5: Calculate moles of acetic acid adsorbed per gram of charcoal.
We have 5 g of activated charcoal, so we can calculate the moles adsorbed per gram:

\[ \text{Moles adsorbed per gram} = \frac{1.0 \times 10^{-3} \, \text{mol}}{5 \, \text{g}} = 2.0 \times 10^{-4} \, \text{mol/g} \]

### Step 6: Calculate the weight of acetic acid adsorbed per gram of charcoal.
To find the weight adsorbed per gram, we multiply the moles adsorbed per gram by the molar mass of acetic acid (which is approximately 60 g/mol):

\[ \text{Weight adsorbed per gram} = \text{Moles adsorbed per gram} \times \text{Molar mass} \]

Calculating:

\[ \text{Weight adsorbed per gram} = 2.0 \times 10^{-4} \, \text{mol/g} \times 60 \, \text{g/mol} = 0.012 \, \text{g} \]

### Final Answer:
The weight of acetic acid adsorbed per gram of charcoal is **0.012 g**.

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