`1.23g` of a substance dissolved in 10g of water raised the boiling point of water to `100.39^(@)C`. Calculate the molecular weight of the substance .`(K_(b)=0.52^(@)Cm^(-1))`

Answer

To calculate the molecular weight of the substance, we can follow these steps:

### Step 1: Identify the given data
- Mass of the solute (substance) = 1.23 g
- Mass of the solvent (water) = 10 g
- Boiling point elevation (ΔT_b) = 100.39 °C - 100 °C = 0.39 °C
- Boiling point elevation constant (K_b) = 0.52 °C kg⁻¹

### Step 2: Convert the mass of water to kilograms
Since molality is defined as moles of solute per kilogram of solvent, we need to convert the mass of water from grams to kilograms:
\[ \text{Mass of water in kg} = \frac{10 \text{ g}}{1000} = 0.01 \text{ kg} \]

### Step 3: Use the formula for boiling point elevation
The relationship between boiling point elevation and molality is given by:
\[ \Delta T_b = K_b \times m \]
Where:
- \( m \) is the molality of the solution.

We can rearrange this formula to find molality:
\[ m = \frac{\Delta T_b}{K_b} \]

### Step 4: Calculate the molality
Substituting the values into the equation:
\[ m = \frac{0.39 \text{ °C}}{0.52 \text{ °C kg}^{-1}} \]
\[ m = 0.75 \text{ mol/kg} \]

### Step 5: Relate molality to moles of solute
Molality (m) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Let \( n \) be the number of moles of the solute. Therefore:
\[ 0.75 = \frac{n}{0.01} \]
\[ n = 0.75 \times 0.01 = 0.0075 \text{ moles} \]

### Step 6: Calculate the molecular weight
The molecular weight (M) can be calculated using the formula:
\[ M = \frac{\text{mass of solute}}{\text{moles of solute}} \]
Substituting the values:
\[ M = \frac{1.23 \text{ g}}{0.0075 \text{ moles}} \]
\[ M = 164 \text{ g/mol} \]

### Final Answer
The molecular weight of the substance is **164 g/mol**.

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