`Delta_(vap)`S1 of acetone is `93.0 JK^(-1) "mol"^(-1)`.If boiling point of acetone is `56^(@)C`, calculate the heat required for the vaporisation of 1 g of acetone. (Answer in Joule)

Explanation

To calculate the heat required for the vaporization of 1 g of acetone, we can follow these steps:

### Step 1: Convert the boiling point of acetone from Celsius to Kelvin.
The boiling point of acetone is given as \( 56^\circ C \). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

Substituting the given value:

\[ T = 56 + 273 = 329 \, K \]

### Step 2: Calculate the enthalpy of vaporization (\( \Delta H_{vap} \)).
We know that the relationship between the enthalpy of vaporization and the entropy of vaporization is given by:

\[ \Delta H_{vap} = T \times \Delta S_{vap} \]

Where:
- \( \Delta S_{vap} = 93.0 \, J \, K^{-1} \, mol^{-1} \)
- \( T = 329 \, K \)

Now, substituting the values:

\[ \Delta H_{vap} = 329 \, K \times 93.0 \, J \, K^{-1} \, mol^{-1} = 30597 \, J \, mol^{-1} \]

### Step 3: Calculate the heat required for vaporizing 1 g of acetone.
The molar mass of acetone (C\(_3\)H\(_6\)O) is approximately 58 g/mol. We need to find the heat required for 1 g of acetone. We can use the following proportion:

\[ \text{Heat required for 1 g} = \frac{\Delta H_{vap}}{\text{Molar mass}} \times \text{mass of acetone} \]

Substituting the values:

\[ \text{Heat required for 1 g} = \frac{30597 \, J \, mol^{-1}}{58 \, g \, mol^{-1}} \times 1 \, g \]

Calculating this gives:

\[ \text{Heat required for 1 g} = \frac{30597}{58} \approx 527.5 \, J \]

### Final Answer:
The heat required for the vaporization of 1 g of acetone is approximately **527.5 Joules**.

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