Statement-1 : Molality of pure ethanol is lesser than pure water
Statement-2 : As density of ethanol is lesser than density of water
[Given : `d_("ethanol") = 0.789 gm//ml, d_("water") = 1gm//ml`]

Options

A. Statement-1 is true, statement -2 is true and statement-2 is correct explanation for statement-1
B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1
C. Statement-1 is false, statement-2 is true
D. Statement-1 is true, statement-2 is false

Answer

To solve the question, we need to analyze both statements and determine their validity based on the definitions and relationships of molality, density, and the properties of pure ethanol and water.

### Step-by-Step Solution:

1. **Understanding Molality**:
- Molality (m) is defined as the number of moles of solute per kilogram of solvent.
- The formula for molality is:
\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]

2. **Density of Ethanol and Water**:
- The density of ethanol is given as \( d_{\text{ethanol}} = 0.789 \, \text{g/ml} \).
- The density of water is given as \( d_{\text{water}} = 1 \, \text{g/ml} \).

3. **Mass Calculation**:
- For 1 ml of each liquid:
- Mass of 1 ml of ethanol = \( 0.789 \, \text{g} \)
- Mass of 1 ml of water = \( 1 \, \text{g} \)

4. **Moles of Solute**:
- Assuming we are comparing the molality of pure ethanol and pure water, we need to consider the mass of the solute (which is 1 ml of each liquid).
- The number of moles of solute can be calculated using the formula:
\[ \text{Moles} = \frac{\text{mass of solute (g)}}{\text{molecular weight (g/mol)}} \]
- The molecular weight of ethanol (C2H5OH) is approximately \( 46 \, \text{g/mol} \) and for water (H2O) it is approximately \( 18 \, \text{g/mol} \).

5. **Calculating Moles**:
- For ethanol:
\[ \text{Moles of ethanol} = \frac{0.789 \, \text{g}}{46 \, \text{g/mol}} \approx 0.0171 \, \text{mol} \]
- For water:
\[ \text{Moles of water} = \frac{1 \, \text{g}}{18 \, \text{g/mol}} \approx 0.0556 \, \text{mol} \]

6. **Calculating Molality**:
- Since we are considering pure liquids, the solvent mass for ethanol is 0.789 g (0.000789 kg) and for water is 1 g (0.001 kg).
- Therefore, the molality of ethanol:
\[ \text{Molality of ethanol} = \frac{0.0171 \, \text{mol}}{0.000789 \, \text{kg}} \approx 21.7 \, \text{mol/kg} \]
- And for water:
\[ \text{Molality of water} = \frac{0.0556 \, \text{mol}}{0.001 \, \text{kg}} \approx 55.6 \, \text{mol/kg} \]

7. **Comparing Molalities**:
- From the calculations, we find that the molality of pure water is greater than that of pure ethanol.
- Thus, **Statement 1** is true: "Molality of pure ethanol is lesser than pure water".

8. **Analyzing Statement 2**:
- Statement 2 claims that "As the density of ethanol is lesser than density of water".
- This statement is also true since \( d_{\text{ethanol}} = 0.789 \, \text{g/ml} < d_{\text{water}} = 1 \, \text{g/ml} \).

9. **Conclusion**:
- Both statements are true, but Statement 2 does not explain Statement 1 correctly. The molality comparison is based on the mass of solute and solvent, not merely on density.

### Final Answer:
- **Statement 1 is true**.
- **Statement 2 is true** but does not correctly explain Statement 1.