Find the elevation in boiling point and (ii) depression in freezing point of a solution containing 0.520g glucose `(C_(6)H_(12)O_(6))` dissolved in 80.2g of water. For water `K_(f)= 1.86k//m. K_(b)=0.52 K//m.`

Explanation

To solve the problem of finding the elevation in boiling point and depression in freezing point of a solution containing 0.520 g of glucose (C₆H₁₂O₆) dissolved in 80.2 g of water, we will follow these steps:

### Step 1: Calculate the molar mass of glucose (C₆H₁₂O₆)

The molar mass of glucose can be calculated as follows:
- Carbon (C): 6 atoms × 12.01 g/mol = 72.06 g/mol
- Hydrogen (H): 12 atoms × 1.008 g/mol = 12.096 g/mol
- Oxygen (O): 6 atoms × 16.00 g/mol = 96.00 g/mol

Adding these together:
\[ \text{Molar mass of glucose} = 72.06 + 12.096 + 96.00 = 180.156 \, \text{g/mol} \]
For simplicity, we can round this to 180 g/mol.

### Step 2: Calculate the number of moles of glucose

Using the formula:
\[ \text{Moles of solute} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \]

Substituting the values:
\[ \text{Moles of glucose} = \frac{0.520 \, \text{g}}{180 \, \text{g/mol}} = 0.00289 \, \text{mol} \]

### Step 3: Calculate the mass of the solvent in kg

Convert the mass of water from grams to kilograms:
\[ \text{Mass of solvent (kg)} = \frac{80.2 \, \text{g}}{1000} = 0.0802 \, \text{kg} \]

### Step 4: Calculate the molality of the solution

Using the formula for molality (m):
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]

Substituting the values:
\[ m = \frac{0.00289 \, \text{mol}}{0.0802 \, \text{kg}} = 0.0360 \, \text{mol/kg} \]

### Step 5: Calculate the elevation in boiling point (ΔT_b)

Using the formula:
\[ \Delta T_b = K_b \times m \]

Substituting the values:
\[ \Delta T_b = 0.52 \, \text{K/m} \times 0.0360 \, \text{mol/kg} = 0.01872 \, \text{K} \]
Rounding this to three significant figures gives:
\[ \Delta T_b \approx 0.019 \, \text{K} \]

### Step 6: Calculate the depression in freezing point (ΔT_f)

Using the formula:
\[ \Delta T_f = K_f \times m \]

Substituting the values:
\[ \Delta T_f = 1.86 \, \text{K/m} \times 0.0360 \, \text{mol/kg} = 0.067056 \, \text{K} \]
Rounding this to three significant figures gives:
\[ \Delta T_f \approx 0.067 \, \text{K} \]

### Final Results:
- Elevation in boiling point (ΔT_b) ≈ 0.019 K
- Depression in freezing point (ΔT_f) ≈ 0.067 K

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