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 can use the formula for heat absorbed (q):
\[ q = m \times c \times \Delta T \]
where:
- \( m \) = mass of water = 200 g
- \( c \) = specific heat capacity of water = 1 cal/g°C
- \( \Delta T \) = change in temperature = 5°C

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

### Step 2: Convert heat from calories to kilocalories
Since \( 1 \, \text{kcal} = 1000 \, \text{cal} \):
\[ q = \frac{1000 \, \text{cal}}{1000} = 1 \, \text{kcal} \]

### Step 3: Calculate the number of moles of butanol
The molar mass of butanol (C4H10O) is approximately 74 g/mol. To find the number of moles of butanol burned:
\[ \text{moles of butanol} = \frac{\text{mass of butanol}}{\text{molar mass of butanol}} = \frac{0.2 \, \text{g}}{74 \, \text{g/mol}} \approx 0.0027027 \, \text{mol} \]

### Step 4: Calculate the heat of combustion per mole of butanol
Since 0.0027027 moles of butanol produced 1 kcal of heat, we can find the heat of combustion per mole:
\[ \text{Heat of combustion} = \frac{1 \, \text{kcal}}{0.0027027 \, \text{mol}} \approx 370 \, \text{kcal/mol} \]

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

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