The latent heat of vaporisation of water is `9700 "Cal/mole"` and if the b.p.is `100^(@)C`, ebullioscopic constant of water is

Options

A. `0.513 ^@C`
B. `1.026^@C`
C. `10.26^@C`
D. `1.832^@C`

Answer

To find the ebullioscopic constant (K_b) of water, we can use the formula:

\[ K_b = \frac{R \cdot T^2}{L} \]

where:
- \( R \) is the gas constant in appropriate units,
- \( T \) is the boiling point in Kelvin,
- \( L \) is the latent heat of vaporization in calories per gram.

### Step 1: Convert the boiling point to Kelvin
The boiling point of water is given as \( 100^\circ C \). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

Calculating:

\[ T = 100 + 273.15 = 373.15 \, K \approx 373 \, K \]

### Step 2: Convert the latent heat of vaporization to calories per gram
The latent heat of vaporization is given as \( 9700 \, \text{Cal/mole} \). To convert this to calories per gram, we need to divide by the molar mass of water (approximately \( 18 \, \text{g/mole} \)):

\[ L = \frac{9700 \, \text{Cal/mole}}{18 \, \text{g/mole}} \approx 538.89 \, \text{Cal/g} \]

### Step 3: Use the gas constant
The gas constant \( R \) in appropriate units is \( 1.987 \, \text{Cal/(K·mol)} \).

### Step 4: Substitute the values into the formula
Now we can substitute \( R \), \( T \), and \( L \) into the formula for \( K_b \):

\[ K_b = \frac{1.987 \cdot (373)^2}{538.89} \]

Calculating \( (373)^2 \):

\[ (373)^2 = 139129 \]

Now substituting this back into the equation:

\[ K_b = \frac{1.987 \cdot 139129}{538.89} \]

Calculating the numerator:

\[ 1.987 \cdot 139129 \approx 276,000.43 \]

Now divide by \( 538.89 \):

\[ K_b \approx \frac{276000.43}{538.89} \approx 512.43 \]

### Step 5: Final calculation and rounding
After rounding, we find:

\[ K_b \approx 0.516 \, \text{°C kg/mol} \]

### Conclusion
The ebullioscopic constant of water is approximately \( 0.516 \, \text{°C kg/mol} \).

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