The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality `("in mol kg"^(-1))` of the aqueous solution is:

Options

A. 13.88
B. `13.88xx10^(-2)`
C. `13.88xx10^(-3)`
D. `13.88xx10^(-1)`

Explanation

To solve the problem of finding the molality of an aqueous solution given the mole fraction of the solvent, we can follow these steps:

### Step 1: Understand the Given Information
We are given the mole fraction of the solvent (water) in an aqueous solution of a solute, which is 0.8.

### Step 2: Calculate the Mole Fraction of the Solute
Since the total mole fraction of the solution is equal to 1, we can find the mole fraction of the solute (A) using the formula:
\[ X_{solute} = 1 - X_{solvent} \]
Substituting the given value:
\[ X_{solute} = 1 - 0.8 = 0.2 \]

### Step 3: Relate Mole Fraction to Moles
Let’s denote:
- \( n_{solvent} \) = moles of solvent (water)
- \( n_{solute} \) = moles of solute

From the mole fractions, we can express the moles of solute and solvent:
\[ X_{solvent} = \frac{n_{solvent}}{n_{solute} + n_{solvent}} \quad \text{and} \quad X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]
Given that \( X_{solvent} = 0.8 \) and \( X_{solute} = 0.2 \), we can set:
\[ n_{solvent} = 0.8n \quad \text{and} \quad n_{solute} = 0.2n \]
where \( n \) is the total number of moles in the solution.

### Step 4: Calculate the Mass of the Solvent
The molar mass of water (H₂O) is approximately 18 g/mol. To find the mass of the solvent in kilograms:
\[ \text{Mass of solvent} = n_{solvent} \times \text{Molar mass of water} \]
Substituting the values:
\[ n_{solvent} = 0.8n \quad \text{and} \quad \text{Molar mass} = 18 \, \text{g/mol} = 0.018 \, \text{kg/mol} \]
Thus,
\[ \text{Mass of solvent} = 0.8n \times 0.018 \, \text{kg} \]

### Step 5: Calculate the Molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ m = \frac{n_{solute}}{\text{Mass of solvent in kg}} \]
Substituting the values:
\[ m = \frac{0.2n}{0.8n \times 0.018} \]
The \( n \) cancels out:
\[ m = \frac{0.2}{0.8 \times 0.018} \]
Calculating the denominator:
\[ 0.8 \times 0.018 = 0.0144 \]
Thus,
\[ m = \frac{0.2}{0.0144} \approx 13.88 \, \text{mol/kg} \]

### Final Answer
The molality of the aqueous solution is approximately **13.88 mol/kg**.

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