1.00g of non - electrolyte dissolved in 100 g of `CS_(2)` , the freezing point lowered by 0.40 K. Find the molar mass of the solute .
(`k_(f)` for `CS_(2)` ` = 5.12 K kg mol^(-1)`).

Explanation

To find the molar mass of the non-electrolyte solute, we can use the formula for freezing point depression:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \(\Delta T_f\) = change in freezing point (0.40 K)
- \(i\) = van 't Hoff factor (for non-electrolytes, \(i = 1\))
- \(K_f\) = freezing point depression constant for \(CS_2\) (5.12 K kg mol\(^{-1}\))
- \(m\) = molality of the solution

### Step 1: Calculate molality (m)
Molality is defined as the number of moles of solute per kilogram of solvent.

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

Given:
- Mass of solute = 1 g
- Mass of solvent (CS\(_2\)) = 100 g = 0.1 kg

To find the number of moles of solute, we use the formula:

\[ \text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{1 \text{ g}}{M} \]

Where \(M\) is the molar mass of the solute.

Substituting this into the molality formula:

\[ m = \frac{1/M}{0.1} = \frac{10}{M} \]

### Step 2: Substitute into the freezing point depression equation
Now we can substitute \(m\) into the freezing point depression equation:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Substituting the known values:

\[ 0.40 = 1 \cdot 5.12 \cdot \frac{10}{M} \]

### Step 3: Solve for M
Now we can solve for \(M\):

\[ 0.40 = \frac{51.2}{M} \]

Multiplying both sides by \(M\):

\[ 0.40M = 51.2 \]

Now, divide both sides by 0.40:

\[ M = \frac{51.2}{0.40} = 128 \text{ g/mol} \]

### Conclusion
The molar mass of the non-electrolyte solute is **128 g/mol**.

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