When 0.2 g of butanol was burnt in a suitable apparatus the heat evolved was sufficient to rise the temperature 200 g of water by `5^(@)C`. The heat of combustion of butanol in kcal/mole will be

Options

A. 1.37
B. 370
C. 3700
D. 307

Answer

To find the heat of combustion of butanol in kcal/mole, we can follow these steps:

### Step 1: Calculate the heat absorbed by water
We know that the heat absorbed by the water can be calculated using the formula:
\[ q = mc\Delta T \]
where:
- \( m \) = mass of water = 200 g
- \( c \) = specific heat of water = 1 cal/g°C
- \( \Delta T \) = change in temperature = 5°C

Substituting the values:
\[ q = 200 \, \text{g} \times 1 \, \text{cal/g°C} \times 5 \, \text{°C} \]
\[ q = 1000 \, \text{cal} \]

### Step 2: Convert calories to kilocalories
Since 1 kcal = 1000 cal, we can convert the heat absorbed by the water to kilocalories:
\[ q = \frac{1000 \, \text{cal}}{1000} = 1 \, \text{kcal} \]

### Step 3: Relate the heat evolved to the amount of butanol burned
The heat evolved by burning 0.2 g of butanol is 1 kcal.

### Step 4: Calculate the heat of combustion per mole of butanol
To find the heat of combustion per mole, we need to determine how much heat would be released by 1 mole of butanol. The molecular weight of butanol (C4H10O) is 74 g/mole.

Using a proportion:
- 0.2 g of butanol releases 1 kcal
- 74 g of butanol releases \( x \) kcal

Setting up the proportion:
\[ x = \frac{74 \, \text{g} \times 1 \, \text{kcal}}{0.2 \, \text{g}} \]
\[ x = \frac{74}{0.2} \, \text{kcal} \]
\[ x = 370 \, \text{kcal} \]

### Step 5: Conclusion
The heat of combustion of butanol is 370 kcal/mole.

### Final Answer
The heat of combustion of butanol is **370 kcal/mole**.

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