Assertion `:-` The molarity of a certain solution is always greater than molority. If density of solution is nearly `1gm//ml`.
Reason `:-` `m=(1000M)/(1000M-MxxM_(solute))`

Options

A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.
B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertiion.
C. If Assertion is True but the Reason is False.
D. If both Assertion & Reason are False

Answer

To solve the given assertion and reason question, we need to establish the relationship between molarity (M) and molality (m) of a solution, especially when the density (D) of the solution is approximately 1 g/mL.

### Step-by-Step Solution:

1. **Understanding Molarity and Molality**:
- Molarity (M) is defined as the number of moles of solute per liter of solution.
- Molality (m) is defined as the number of moles of solute per kilogram of solvent.

2. **Formulas**:
- Molarity (M) can be expressed as:
\[ M = \frac{n}{V} = \frac{W/M_m}{V} \]
where \( n \) is the number of moles, \( W \) is the mass of the solute, \( M_m \) is the molar mass of the solute, and \( V \) is the volume of the solution in liters.
- Molality (m) can be expressed as:
\[ m = \frac{n}{W_{solvent}} = \frac{W/M_m}{W_{solvent}} \]
where \( W_{solvent} \) is the mass of the solvent in kilograms.

3. **Relating Molarity and Molality**:
- The density (D) of the solution is defined as:
\[ D = \frac{W_{solution}}{V} \]
where \( W_{solution} = W + W_{solvent} \).
- If we assume the density is approximately 1 g/mL, then 1 L of solution weighs about 1000 g.

4. **Substituting Values**:
- From the density equation, we can express the volume \( V \) in terms of mass:
\[ V = \frac{W + W_{solvent}}{D} \]
- If \( D \approx 1 \, \text{g/mL} \), then \( V \approx W + W_{solvent} \).

5. **Finding the Relationship**:
- Substitute \( V \) into the molarity equation:
\[ M = \frac{W/M_m}{(W + W_{solvent})/1000} \]
- Rearranging gives:
\[ M = \frac{1000W}{M_m(W + W_{solvent})} \]
- For molality:
\[ m = \frac{W/M_m}{W_{solvent}/1000} \]
- Rearranging gives:
\[ m = \frac{1000W}{M_m W_{solvent}} \]

6. **Final Relationship**:
- By substituting \( W_{solvent} \) in terms of molality and molarity, we can derive:
\[ m = \frac{1000M}{1000 - M \cdot M_m} \]
- This shows that molarity is indeed greater than molality when density is approximately 1 g/mL.

### Conclusion:
- Since the derived relationship shows that molarity (M) is greater than molality (m) under the given conditions, both the assertion and the reason are correct.