What will be the temperature at which a solution containing 6 g of glucose per 1000 g water will boil if molal elevation constant for what is 0.52 K kg `mol^(-1)`.

Options

A. `1000.173^@C`
B. `100.0173^@C`
C. `100.173^@C`
D. None

Answer

To solve the problem, we need to calculate the boiling point elevation of the solution containing glucose in water. Here are the steps to find the boiling point of the solution:

### Step 1: Calculate the number of moles of glucose
We start by calculating the number of moles of glucose (C6H12O6) present in the solution.

1. **Molar mass of glucose**:
- C: 12 g/mol × 6 = 72 g/mol
- H: 1 g/mol × 12 = 12 g/mol
- O: 16 g/mol × 6 = 96 g/mol
- Total = 72 + 12 + 96 = 180 g/mol

2. **Mass of glucose**: 6 g

3. **Number of moles of glucose**:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{6 \text{ g}}{180 \text{ g/mol}} = 0.0333 \text{ mol} \]

### Step 2: Calculate the mass of the solvent (water)
The mass of the solvent (water) is given as 1000 g.

### Step 3: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

1. **Mass of solvent in kg**:
\[ 1000 \text{ g} = 1 \text{ kg} \]

2. **Molality (m)**:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0333 \text{ mol}}{1 \text{ kg}} = 0.0333 \text{ mol/kg} \]

### Step 4: Calculate the boiling point elevation
The formula for boiling point elevation is given by:
\[ \Delta T_b = i \cdot m \cdot K_b \]
Where:
- \( \Delta T_b \) = boiling point elevation
- \( i \) = van 't Hoff factor (for glucose, which is a non-electrolyte, \( i = 1 \))
- \( K_b \) = molal boiling point elevation constant (given as 0.52 K kg/mol)

Substituting the values:
\[ \Delta T_b = 1 \cdot 0.0333 \text{ mol/kg} \cdot 0.52 \text{ K kg/mol} = 0.0173 \text{ K} \]

### Step 5: Calculate the new boiling point of the solution
The normal boiling point of water is 100 °C. The new boiling point of the solution is:
\[ \text{Boiling point of solution} = \text{Boiling point of solvent} + \Delta T_b \]
\[ \text{Boiling point of solution} = 100 \text{ °C} + 0.0173 \text{ °C} = 100.0173 \text{ °C} \]

### Final Answer
The boiling point of the solution is approximately **100.0173 °C**.

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