6g of non-volatile, non-electrolyte `x` dissolved in 100g of water freezes `-0.93^(@)C`. The molar mass of `x` in `"g mol"^(-1)` is :
`(K_(f) "of" H_(2)O=1.86 "K kg mol"^(-1))`

Options

A. 60
B. 140
C. 180
D. 120

Explanation

To find the molar mass of the non-volatile, non-electrolyte solute \( x \) that is dissolved in water, we can use the freezing point depression formula. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We are given:
- Mass of solute \( x = 6 \, \text{g} \)
- Mass of solvent (water) = \( 100 \, \text{g} = 0.1 \, \text{kg} \)
- Freezing point depression \( \Delta T_f = -0.93 \, ^\circ C \)
- Freezing point depression constant for water \( K_f = 1.86 \, \text{K kg mol}^{-1} \)

### Step 2: Calculate the Freezing Point Depression
The formula for freezing point depression is:
\[ \Delta T_f = K_f \times m \]
where \( m \) is the molality of the solution.

### Step 3: Calculate Molality
Molality \( m \) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
First, we need to calculate the number of moles of solute \( x \):
\[ \text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{6 \, \text{g}}{M} \]
where \( M \) is the molar mass of solute \( x \).

Now substituting this into the molality formula:
\[ m = \frac{6/M}{0.1} = \frac{60}{M} \]

### Step 4: Substitute into the Freezing Point Depression Formula
Now we can substitute \( m \) into the freezing point depression formula:
\[ 0.93 = 1.86 \times \frac{60}{M} \]

### Step 5: Solve for Molar Mass \( M \)
Rearranging the equation to solve for \( M \):
\[ M = 1.86 \times \frac{60}{0.93} \]

Calculating the right side:
\[ M = \frac{111.6}{0.93} \approx 120 \, \text{g/mol} \]

### Conclusion
The molar mass of the solute \( x \) is approximately \( 120 \, \text{g/mol} \).