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`
What is the quantity of water that should be added to 16 g methonal to make the mole fraction of methonal as 0.25?

Options

A. 27 gm
B. 12gm
C. 18 gm
D. 36 gm

Explanation

To solve the problem of determining the quantity of water that should be added to 16 g of methanol to achieve a mole fraction of methanol (X_A) of 0.25, we can follow these steps:

### Step 1: Identify the known values
- Weight of methanol (solute), \( w_A = 16 \, \text{g} \)
- Molar mass of methanol, \( m_A = 32 \, \text{g/mol} \)
- Desired mole fraction of methanol, \( X_A = 0.25 \)
- Molar mass of water (solvent), \( m_B = 18 \, \text{g/mol} \)

### Step 2: Use the mole fraction formula
The mole fraction of methanol can be expressed as:
\[ X_A = \frac{n}{N + n} \]
Where:
- \( n \) = moles of methanol
- \( N \) = moles of water

### Step 3: Calculate the moles of methanol
Using the formula for moles:
\[ n = \frac{w_A}{m_A} = \frac{16 \, \text{g}}{32 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 4: Express the mole fraction in terms of moles of water
From the mole fraction formula, we can rearrange it to find \( N \):
\[ X_A = \frac{n}{N + n} \implies 0.25 = \frac{0.5}{N + 0.5} \]

### Step 5: Solve for moles of water (N)
Cross-multiplying gives:
\[ 0.25(N + 0.5) = 0.5 \]
Expanding and rearranging:
\[ 0.25N + 0.125 = 0.5 \implies 0.25N = 0.375 \implies N = \frac{0.375}{0.25} = 1.5 \, \text{mol} \]

### Step 6: Calculate the weight of water (w_B)
Now, we convert moles of water to grams:
\[ w_B = N \times m_B = 1.5 \, \text{mol} \times 18 \, \text{g/mol} = 27 \, \text{g} \]

### Final Answer
The quantity of water that should be added is **27 grams**.

---