When 45g of glucose was dissolved in `500g` of water the solution has a freezing point of `-0.93^(@)C`.
(a) What is the molecular weight of glucose ?
(b) If the simplest formula is `CH_(2)O` what is its molecular formula ?

Answer

To solve the problem step by step, we will first calculate the molecular weight of glucose using the freezing point depression formula, and then we will determine its molecular formula based on the empirical formula provided.

### Step-by-Step Solution

**(a) Finding the Molecular Weight of Glucose:**

1. **Identify Given Values:**
- Mass of glucose (solute) = 45 g
- Mass of water (solvent) = 500 g
- Depression in freezing point (ΔTf) = -0.93 °C
- Freezing point depression constant (Kf) for water = 1.86 °C/m

2. **Use the Freezing Point Depression Formula:**
\[ \Delta T_f = K_f \cdot m \]
where \( m \) is the molality of the solution.

3. **Calculate Molality (m):**
- First, convert the mass of the solvent (water) from grams to kilograms:
\[ \text{Mass of water} = 500 \, \text{g} = 0.500 \, \text{kg} \]
- Rearranging the formula for molality:
\[ m = \frac{\Delta T_f}{K_f} \]
- Substitute the values:
\[ m = \frac{0.93}{1.86} = 0.5 \, \text{mol/kg} \]

4. **Calculate Moles of Glucose:**
- Using the definition of molality:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
- Rearranging gives:
\[ \text{moles of glucose} = m \cdot \text{mass of solvent in kg} = 0.5 \cdot 0.500 = 0.25 \, \text{moles} \]

5. **Calculate the Molar Mass of Glucose:**
- Molar mass (M) can be calculated using the formula:
\[ M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{45 \, \text{g}}{0.25 \, \text{moles}} = 180 \, \text{g/mol} \]

**(b) Finding the Molecular Formula:**

1. **Identify the Empirical Formula:**
- Given empirical formula = CH₂O

2. **Calculate the Molar Mass of the Empirical Formula:**
- Molar mass of CH₂O:
\[ C: 12 \, \text{g/mol} + H: 2 \cdot 1 \, \text{g/mol} + O: 16 \, \text{g/mol} = 30 \, \text{g/mol} \]

3. **Determine the Ratio (n) of Molecular Mass to Empirical Mass:**
- Using the molar mass of glucose (180 g/mol) and the empirical formula mass (30 g/mol):
\[ n = \frac{\text{Molar mass of glucose}}{\text{Empirical formula mass}} = \frac{180}{30} = 6 \]

4. **Calculate the Molecular Formula:**
- Multiply the subscripts in the empirical formula by n:
\[ \text{Molecular formula} = (CH_2O)_6 = C_6H_{12}O_6 \]

### Final Answers:
- (a) The molecular weight of glucose is **180 g/mol**.
- (b) The molecular formula of glucose is **C₆H₁₂O₆**.