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

Options

A. 0.25
B. 0.66
C. 0.33
D. 0.75

Answer

To solve the problem, we need to find the ratio of the molar masses of solutes A and B based on the given information about their osmotic pressures.

### Step-by-Step Solution:

1. **Understanding the Osmotic Pressure Formula**:
The osmotic pressure (π) is given by the formula:
\[ \pi = iCRT \]
where:
- \( i \) = van 't Hoff factor (which is 1 for non-electrolytes),
- \( C \) = concentration of the solution (in moles per liter),
- \( R \) = gas constant,
- \( T \) = temperature (in Kelvin).

Since \( i \), \( R \), and \( T \) are constant for both solutions, we can say that the osmotic pressure is directly proportional to the concentration.

2. **Setting Up the Moles for Each Solution**:
For the first solution:
- Mass of solute A = 10 g
- Mass of solute B = 20 g
- Moles of A = \( \frac{10}{M_A} \)
- Moles of B = \( \frac{20}{M_B} \)

Therefore, the total moles for the first solution:
\[ \text{Total moles (Solution 1)} = \frac{10}{M_A} + \frac{20}{M_B} \]

For the second solution:
- Mass of solute A = 6.67 g
- Mass of solute B = 30 g
- Moles of A = \( \frac{6.67}{M_A} \)
- Moles of B = \( \frac{30}{M_B} \)

Therefore, the total moles for the second solution:
\[ \text{Total moles (Solution 2)} = \frac{6.67}{M_A} + \frac{30}{M_B} \]

3. **Equating the Total Moles**:
Since both solutions have the same osmotic pressure, we can set the total moles equal to each other:
\[ \frac{10}{M_A} + \frac{20}{M_B} = \frac{6.67}{M_A} + \frac{30}{M_B} \]

4. **Rearranging the Equation**:
To eliminate the fractions, we can multiply through by \( M_A \cdot M_B \):
\[ 10M_B + 20M_A = 6.67M_B + 30M_A \]

Rearranging gives:
\[ 10M_B - 6.67M_B = 30M_A - 20M_A \]
\[ 3.33M_B = 10M_A \]

5. **Finding the Ratio of Molar Masses**:
Now, we can express the ratio of the molar masses:
\[ \frac{M_A}{M_B} = \frac{3.33}{10} = 0.333 \]

Therefore, the ratio of the molar masses of A and B is:
\[ \frac{M_A}{M_B} = 0.33 \]

### Final Answer:
The ratio of the molar masses of A and B is \( \frac{M_A}{M_B} = 0.33 \).

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