Elevation in boiling point was `0.52^(@)C` when 6 g of a compound X was dissolved in 100 g of water. Molecular weight of X is (`K_(b)` of water is `5.2^(@)C` per 100 g of water)

Options

A. 120
B. 60
C. 180
D. 342

Answer

To find the molecular weight of compound X based on the given elevation in boiling point, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Elevation in boiling point (ΔTb) = 0.52°C
- Kb (boiling point elevation constant for water) = 5.2°C per 100 g of water
- Mass of solute (compound X) = 6 g
- Mass of solvent (water) = 100 g

2. **Convert Kb to per gram:**
- Since Kb is given for 100 g of water, we need to convert it to per gram:
\[ Kb = \frac{5.2°C}{100 \text{ g}} = 0.052°C/\text{g} \]

3. **Use the Formula for Elevation in Boiling Point:**
The formula for elevation in boiling point is:
\[ \Delta Tb = Kb \cdot m \]
where \( m \) is the molality.

4. **Calculate Molality (m):**
Molality (m) is defined as:
\[ m = \frac{\text{mass of solute (g)} \times 1000}{\text{molar mass of solute (g/mol)} \times \text{mass of solvent (g)}} \]
Rearranging the formula gives:
\[ m = \frac{6 \text{ g} \times 1000}{M \times 100 \text{ g}} \]
where \( M \) is the molar mass of the solute (X).

5. **Substituting Values into the Elevation Formula:**
Now substituting the values into the elevation formula:
\[ 0.52 = 0.052 \cdot \left(\frac{6 \times 1000}{M \times 100}\right) \]

6. **Simplifying the Equation:**
Simplifying the equation:
\[ 0.52 = 0.052 \cdot \left(\frac{6000}{M}\right) \]
\[ 0.52 = \frac{312}{M} \]

7. **Solving for Molar Mass (M):**
Rearranging gives:
\[ M = \frac{312}{0.52} \]
\[ M = 600 \text{ g/mol} \]

### Conclusion:
The molecular weight of compound X is 600 g/mol.