The relative lowering of vapour pressure of an aqueous solution of urea is 0.018. If `K_(b)` for `H_(2)O` is `0.54^(@)m^(-1)`, the elevation in boiling point will be

Options

A. `0.54^(@)C`
B. `0.18^(@)C`
C. 0.54 K
D. 0.18 K

Answer

To solve the problem of finding the elevation in boiling point of an aqueous solution of urea given the relative lowering of vapor pressure, we can follow these steps:

### Step 1: Understand the Relative Lowering of Vapor Pressure
The relative lowering of vapor pressure (ΔP/P₀) is given as 0.018. This can be expressed in terms of the mole fraction of the solute (urea) in the solution:
\[ \frac{\Delta P}{P_0} = X_{\text{urea}} = 0.018 \]

### Step 2: Calculate the Mole Fraction of Water
The mole fraction of water (X_water) can be calculated as:
\[ X_{\text{water}} = 1 - X_{\text{urea}} = 1 - 0.018 = 0.982 \]

### Step 3: Relate Mole Fraction to Molality
The mole fraction of urea can also be related to molality (m) using the formula:
\[ X_{\text{urea}} = \frac{n_{\text{urea}}}{n_{\text{urea}} + n_{\text{water}}} \]
Where \( n_{\text{urea}} \) is the number of moles of urea and \( n_{\text{water}} \) is the number of moles of water.

Assuming we have 1 kg of water (which is approximately 55.5 moles), we can express:
\[ X_{\text{urea}} = \frac{m_{\text{urea}}}{m_{\text{urea}} + 55.5} \]
Given that \( X_{\text{urea}} = 0.018 \), we can rearrange to find \( m_{\text{urea}} \):
\[ 0.018 = \frac{m_{\text{urea}}}{m_{\text{urea}} + 55.5} \]
Cross-multiplying gives:
\[ 0.018(m_{\text{urea}} + 55.5) = m_{\text{urea}} \]
\[ 0.018m_{\text{urea}} + 1 = m_{\text{urea}} \]
\[ 0.982m_{\text{urea}} = 1 \]
\[ m_{\text{urea}} = \frac{1}{0.982} \approx 1.018 \text{ moles} \]

### Step 4: Calculate Molality
Since we assumed 1 kg of water, the molality (m) is approximately equal to the number of moles of solute (urea) per kg of solvent (water):
\[ m = 1.018 \text{ mol/kg} \]

### Step 5: Use the Elevation in Boiling Point Formula
The elevation in boiling point (ΔT_b) can be calculated using the formula:
\[ \Delta T_b = K_b \times m \]
Where \( K_b \) for water is given as 0.54 °C/m. Thus:
\[ \Delta T_b = 0.54 \, \text{°C/m} \times 1.018 \, \text{m} \approx 0.55 \, \text{°C} \]

### Final Answer
The elevation in boiling point is approximately 0.55 °C.

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