An aqueous glucose solution has boiling point of `374.19K`. Another aqueous glucose solution has a freezing point of `272.22K`. Find the molality ratio of two solution. Boiling point and freezing point of `H_(2)O373.15K` and `273.15K` respectively. `Kb` and `Kf` for water is 0.512 and 1.86 respectively.

Explanation

To find the molality ratio of the two glucose solutions based on their boiling and freezing points, we can follow these steps:

### Step 1: Calculate the change in boiling point (ΔTb) for the first solution.
The boiling point of the first solution is given as 374.19 K, and the boiling point of pure water is 373.15 K.

\[ \Delta T_b = T_{b, solution} - T_{b, solvent} = 374.19 \, K - 373.15 \, K = 1.04 \, K \]

### Step 2: Use the boiling point elevation formula to find molality (M1).
The formula for boiling point elevation is:

\[ \Delta T_b = K_b \cdot m_1 \]

Where:
- \( K_b = 0.512 \, K \cdot kg/mol \) (given)
- \( m_1 \) is the molality of the first solution.

Substituting the values:

\[ 1.04 = 0.512 \cdot m_1 \]

Now, solve for \( m_1 \):

\[ m_1 = \frac{1.04}{0.512} \approx 2.03125 \, mol/kg \]

### Step 3: Calculate the change in freezing point (ΔTf) for the second solution.
The freezing point of the second solution is given as 272.22 K, and the freezing point of pure water is 273.15 K.

\[ \Delta T_f = T_{f, solvent} - T_{f, solution} = 273.15 \, K - 272.22 \, K = 0.93 \, K \]

### Step 4: Use the freezing point depression formula to find molality (M2).
The formula for freezing point depression is:

\[ \Delta T_f = K_f \cdot m_2 \]

Where:
- \( K_f = 1.86 \, K \cdot kg/mol \) (given)
- \( m_2 \) is the molality of the second solution.

Substituting the values:

\[ 0.93 = 1.86 \cdot m_2 \]

Now, solve for \( m_2 \):

\[ m_2 = \frac{0.93}{1.86} \approx 0.5 \, mol/kg \]

### Step 5: Calculate the molality ratio (M1/M2).
Now we can find the molality ratio of the two solutions:

\[ \frac{m_1}{m_2} = \frac{2.03125}{0.5} \approx 4.0625 \]

### Final Result:
The molality ratio of the two glucose solutions is approximately 4.

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