Elevation of boiling point of 1 molar aqueous glucose solution `("density"=1.2g//ml)` is

Options

A. `"K"_("b")`
B. 1.20 `"K"_("b")`
C. 1.02`"K"_("b")`
D. 0.98`"K"_("b")`

Explanation

To solve the problem of finding the elevation of the boiling point of a 1 molar aqueous glucose solution with a density of 1.2 g/ml, we can follow these steps:

### Step 1: Determine the number of moles of glucose
Since the solution is 1 molar, it means there is 1 mole of glucose in 1 liter (1000 ml) of solution.

**Hint:** Molarity (M) is defined as moles of solute per liter of solution.

### Step 2: Calculate the mass of glucose
The molar mass of glucose (C6H12O6) is approximately 180 g/mol. Therefore, the mass of 1 mole of glucose is:
\[ \text{Mass of glucose} = 1 \, \text{mol} \times 180 \, \text{g/mol} = 180 \, \text{g} \]

**Hint:** To find the mass of a substance, multiply the number of moles by its molar mass.

### Step 3: Calculate the total mass of the solution
Using the density of the solution (1.2 g/ml), we can find the total mass of 1 liter (1000 ml) of the solution:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.2 \, \text{g/ml} \times 1000 \, \text{ml} = 1200 \, \text{g} \]

**Hint:** The mass of a solution can be found by multiplying its density by its volume.

### Step 4: Calculate the mass of the solvent (water)
To find the mass of the solvent (water), we subtract the mass of the solute (glucose) from the total mass of the solution:
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} = 1200 \, \text{g} - 180 \, \text{g} = 1020 \, \text{g} \]

**Hint:** The mass of the solvent is the total mass of the solution minus the mass of the solute.

### Step 5: Convert the mass of the solvent to kilograms
Since molality is defined in terms of kilograms of solvent, we convert the mass of the solvent from grams to kilograms:
\[ \text{Mass of solvent in kg} = \frac{1020 \, \text{g}}{1000} = 1.02 \, \text{kg} \]

**Hint:** To convert grams to kilograms, divide by 1000.

### Step 6: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{1 \, \text{mol}}{1.02 \, \text{kg}} \approx 0.9804 \, \text{mol/kg} \]

**Hint:** Molality is calculated using the formula: m = moles of solute / kg of solvent.

### Step 7: Calculate the elevation of boiling point (ΔTb)
The elevation of boiling point can be calculated using the formula:
\[ \Delta T_b = K_b \times m \]
Where \( K_b \) is the ebullioscopic constant for water (approximately 0.512 °C kg/mol). Substituting the values:
\[ \Delta T_b = 0.512 \, \text{°C kg/mol} \times 0.9804 \, \text{mol/kg} \approx 0.501 \, \text{°C} \]

**Hint:** The elevation in boiling point is found by multiplying the ebullioscopic constant by the molality of the solution.

### Final Answer
The elevation of the boiling point of a 1 molar aqueous glucose solution is approximately \( 0.501 \, \text{°C} \).