Molality : It is defined as the moles of the solute pressent in 1 kg of the solvent . It is denoted by m.
Molality(m)`=("Number of moles of solute")/("Number of kilograms of the solvent")`
let `w_(A)` grams of the solute of molecular mass `m_(A)` be present in `w_(B)` grams of the solvent, then:
Molality(m) =`(w_(A))/(m_(A)xxw_(B))xx1000`
Relation between mole fraction and molality:
`X_(A)=(n)/(N+n) "and" X_(B)=(N)/(N+n)`
`(X_(A))/(X_(B))=(n)/(N)=("Moles of solute")/("Moles of solvent")=(w_(A)xxm_(B))/(w_(B)xxm_(A))`
`(X_(A)xx1000)/(X_(B)xxm_(B))=(w_(A)xx1000)/(w_(B)xxm_(A))= m or (X_(A)xx1000)/((1-X_(A))m_(B))=m`
If the mole fraction of a solute is changed from `(1)/(4) "to" (1)/(2)` in the 800 g of solvent then the ratio tof molality will be:

Options

A. `1:3`
B. `1:4`
C. `2:3`
D. `1:2`

Explanation

To solve the problem, we need to calculate the molality of a solution when the mole fraction of a solute changes from \( \frac{1}{4} \) to \( \frac{1}{2} \) in 800 grams of solvent. We'll use the relationship between molality and mole fraction given in the question.

### Step-by-Step Solution:

1. **Understand the Definitions**:
- Molality (\( m \)) is defined as the number of moles of solute per kilogram of solvent.
- The mole fraction of solute (\( X_A \)) and solvent (\( X_B \)) are defined as:
\[ X_A = \frac{n}{N+n} \quad \text{and} \quad X_B = \frac{N}{N+n} \]
where \( n \) is the number of moles of solute and \( N \) is the number of moles of solvent.

2. **Calculate Molality for the First Condition**:
- For the first condition where \( X_A = \frac{1}{4} \):
\[ m = \frac{X_A \times 1000}{(1 - X_A) \times m_B} \]
- Substitute \( X_A = \frac{1}{4} \):
\[ m_1 = \frac{\frac{1}{4} \times 1000}{(1 - \frac{1}{4}) \times m_B} = \frac{250}{\frac{3}{4} \times m_B} = \frac{250 \times 4}{3 \times m_B} = \frac{1000}{3 m_B} \]

3. **Calculate Molality for the Second Condition**:
- For the second condition where \( X_A = \frac{1}{2} \):
\[ m_2 = \frac{X_A \times 1000}{(1 - X_A) \times m_B} \]
- Substitute \( X_A = \frac{1}{2} \):
\[ m_2 = \frac{\frac{1}{2} \times 1000}{(1 - \frac{1}{2}) \times m_B} = \frac{500}{\frac{1}{2} \times m_B} = \frac{500 \times 2}{m_B} = \frac{1000}{m_B} \]

4. **Calculate the Ratio of Molality**:
- Now, we find the ratio of the two molalities:
\[ \text{Ratio} = \frac{m_1}{m_2} = \frac{\frac{1000}{3 m_B}}{\frac{1000}{m_B}} = \frac{1000}{3 m_B} \times \frac{m_B}{1000} = \frac{1}{3} \]

### Final Answer:
The ratio of molality when the mole fraction of the solute changes from \( \frac{1}{4} \) to \( \frac{1}{2} \) is \( \frac{1}{3} \).

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