`10gm` of solute `A` and `20gm` of solute `B` both are dissolved in `500ml`. Of water. The solution ha the same osmotic pressure as `6.67 gm` of `A`30g of `B` are dissolved in the same volume of water at the same temperature. What is the ratio of molar masses of `A` and `B`.

Answer

To solve the problem, we need to find the ratio of the molar masses of solute A and solute B based on the given conditions regarding osmotic pressure.

### Step-by-Step Solution:

1. **Understand the relationship of osmotic pressure**:
The osmotic pressure (\( \Pi \)) of a solution can be expressed using the formula:
\[ \Pi = CRT \]
where \( C \) is the molarity of the solution, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. Since the temperature and \( R \) are constant for both solutions, we can equate the molarities of the two solutions.

2. **Set up the molarity equations**:
For the first solution (10 g of A and 20 g of B):
- Molarity of A (\( C_A \)):
\[ C_A = \frac{10 \, \text{g}}{M_A \times 0.5 \, \text{L}} = \frac{20}{M_A} \]
- Molarity of B (\( C_B \)):
\[ C_B = \frac{20 \, \text{g}}{M_B \times 0.5 \, \text{L}} = \frac{40}{M_B} \]
- Total molarity (\( C_1 \)):
\[ C_1 = C_A + C_B = \frac{20}{M_A} + \frac{40}{M_B} \]

For the second solution (6.67 g of A and 30 g of B):
- Molarity of A (\( C_A' \)):
\[ C_A' = \frac{6.67 \, \text{g}}{M_A \times 0.5 \, \text{L}} = \frac{13.34}{M_A} \]
- Molarity of B (\( C_B' \)):
\[ C_B' = \frac{30 \, \text{g}}{M_B \times 0.5 \, \text{L}} = \frac{60}{M_B} \]
- Total molarity (\( C_2 \)):
\[ C_2 = C_A' + C_B' = \frac{13.34}{M_A} + \frac{60}{M_B} \]

3. **Equate the total molarities**:
Since the osmotic pressures are the same, we have:
\[ C_1 = C_2 \]
Therefore,
\[ \frac{20}{M_A} + \frac{40}{M_B} = \frac{13.34}{M_A} + \frac{60}{M_B} \]

4. **Rearranging the equation**:
To solve for the molar masses, rearranging gives:
\[ \frac{20}{M_A} - \frac{13.34}{M_A} = \frac{60}{M_B} - \frac{40}{M_B} \]
Simplifying this:
\[ \frac{6.66}{M_A} = \frac{20}{M_B} \]

5. **Cross-multiplying**:
Cross-multiplying gives:
\[ 6.66 M_B = 20 M_A \]

6. **Finding the ratio of molar masses**:
Rearranging gives:
\[ \frac{M_A}{M_B} = \frac{6.66}{20} = \frac{1}{3} \]

### Final Result:
The ratio of the molar masses of solute A to solute B is:
\[ \frac{M_A}{M_B} = \frac{1}{3} \]