The relative decrease in `VP` of an aqueous glucose dilute solution is found to be `0.018`. Hence, the elevation in boiling point is (it is given `1 molal` aqueous urea solution boils at `100.54^@)C` at `1 atm` pressure)

Options

A. `0.018^(@)`
B. `0.18^(@)`
C. `0.54^(@)`
D. `0.03^(@)`

Explanation

To solve the problem, we need to find the elevation in boiling point of an aqueous glucose solution given the relative decrease in vapor pressure. Here’s a step-by-step solution:

### Step 1: Understand the relationship between vapor pressure and mole fraction
The relative decrease in vapor pressure (ΔP) is given by the formula:
\[ \frac{P^0 - P}{P^0} = \chi_{solute} \]
where \(P^0\) is the vapor pressure of the pure solvent, \(P\) is the vapor pressure of the solution, and \(\chi_{solute}\) is the mole fraction of the solute.

### Step 2: Calculate the mole fraction of solute (glucose)
Given that the relative decrease in vapor pressure is 0.018, we can equate this to the mole fraction of glucose:
\[ \chi_{glucose} = 0.018 \]

### Step 3: Relate mole fraction to moles of solute and solvent
Let \(n_{solute}\) be the moles of glucose and \(n_{solvent}\) be the moles of water. The mole fraction of glucose can be expressed as:
\[ \chi_{glucose} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]
Assuming \(n_{solute} = 0.018\), we can rearrange the equation to find \(n_{solvent}\):
\[ 0.018 = \frac{0.018}{0.018 + n_{solvent}} \]
Cross-multiplying gives:
\[ 0.018 + n_{solvent} = 0.018 \implies n_{solvent} = 0.982 \]

### Step 4: Calculate the weight of the solvent (water)
The number of moles of water can be converted to grams using the molar mass of water (18 g/mol):
\[ \text{Weight of water} = n_{solvent} \times 18 \text{ g/mol} = 0.982 \times 18 = 17.676 \text{ g} \]

### Step 5: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ m = \frac{n_{solute}}{\text{Weight of solvent in kg}} = \frac{0.018}{17.676 \times 10^{-3}} \approx 1.018 \text{ mol/kg} \]

### Step 6: Use the boiling point elevation formula
The boiling point elevation (\( \Delta T_b \)) can be calculated using the formula:
\[ \Delta T_b = K_b \times m \]
where \(K_b\) is the ebullioscopic constant. However, we can use the boiling point elevation of a 1 molal urea solution as a reference:
Given that a 1 molal urea solution boils at \(100.54^\circ C\), we know:
\[ \Delta T_b \text{ for urea} = 0.54^\circ C \]

### Step 7: Calculate the elevation in boiling point for glucose
Since the molality of glucose is \(1.018\) (which is close to \(1\)), we can find the new boiling point elevation:
\[ \Delta T_b = 0.54 \times 1.018 \approx 0.5498^\circ C \]
Thus, the boiling point of the glucose solution will be:
\[ 100 + 0.5498 \approx 100.55^\circ C \]

### Final Answer
The elevation in boiling point of the aqueous glucose solution is approximately \(0.55^\circ C\).

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