Two solutions of non-volatile and non-electrolyte solute `A` and `B` are prepared separately. The molar mass ratio `(M_(A))/(M_(B))=1/3`. Both are prepared as `5%` by weight solution in water. Then what is the ratio of freezing point depresions, `((DeltaT_(f))_(A))/((DeltaT_(f))_(B))` of the solutions?

Explanation

To solve the problem, we need to find the ratio of freezing point depressions \((\Delta T_f)_A\) and \((\Delta T_f)_B\) for two solutions of non-volatile and non-electrolyte solutes A and B, given that the molar mass ratio \(\frac{M_A}{M_B} = \frac{1}{3}\) and both solutions are 5% by weight in water.

### Step-by-Step Solution:

1. **Understand the Formula for Freezing Point Depression**:
The freezing point depression (\(\Delta T_f\)) is given by the formula:
\[ \Delta T_f = K_f \times m \]
where \(K_f\) is the freezing point depression constant and \(m\) is the molality of the solution.

2. **Calculate the Mass of Solute and Solvent**:
Assume we have 100 g of solution. Since it is a 5% by weight solution:
- Mass of solute (A or B) = 5 g
- Mass of solvent (water) = 100 g - 5 g = 95 g

3. **Convert Mass of Solvent to kg**:
To calculate molality, we need the mass of the solvent in kg:
\[ \text{Mass of solvent} = 95 \, \text{g} = 0.095 \, \text{kg} \]

4. **Calculate the Moles of Solute**:
The number of moles of solute can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{mass of solute}}{\text{molar mass}} \]
For solute A:
\[ n_A = \frac{5 \, \text{g}}{M_A} \]
For solute B:
\[ n_B = \frac{5 \, \text{g}}{M_B} \]

5. **Calculate the Molality of Each Solution**:
The molality \(m\) is given by:
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} \]
For solute A:
\[ m_A = \frac{n_A}{0.095} = \frac{5/M_A}{0.095} = \frac{5}{0.095 M_A} \]
For solute B:
\[ m_B = \frac{n_B}{0.095} = \frac{5/M_B}{0.095} = \frac{5}{0.095 M_B} \]

6. **Substitute into the Freezing Point Depression Formula**:
Now we can express \(\Delta T_f\) for both solutes:
\[ \Delta T_f(A) = K_f \cdot m_A = K_f \cdot \frac{5}{0.095 M_A} \]
\[ \Delta T_f(B) = K_f \cdot m_B = K_f \cdot \frac{5}{0.095 M_B} \]

7. **Find the Ratio of Freezing Point Depressions**:
Now, we can find the ratio:
\[ \frac{\Delta T_f(A)}{\Delta T_f(B)} = \frac{K_f \cdot \frac{5}{0.095 M_A}}{K_f \cdot \frac{5}{0.095 M_B}} = \frac{M_B}{M_A} \]

8. **Use the Given Molar Mass Ratio**:
Given that \(\frac{M_A}{M_B} = \frac{1}{3}\), we can express this as:
\[ \frac{M_B}{M_A} = 3 \]

9. **Final Result**:
Therefore, the ratio of freezing point depressions is:
\[ \frac{\Delta T_f(A)}{\Delta T_f(B)} = 3 \]

### Final Answer:
\[ \frac{\Delta T_f(A)}{\Delta T_f(B)} = 3 \]