1.00 gm of a non-electrolyte solute dissolved in 50 gm of benzene lowered the freezing point of benzene by 0.40 K. `K_(f)` for benzene is 5.12 kg `mol^(-1)`. Molecular mass of the solute will be

Options

A. `256 g mol^(-1)`
B. `2.56 g mol^(-1)`
C. `512xx10^(3)g mol^(-1)`
D. `2.56xx10^(4)g mol^(-1)`

Explanation

To find the molecular mass of the solute, we can use the formula for depression in freezing point:

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

Where:
- \(\Delta T_f\) = depression in freezing point (in Kelvin)
- \(K_f\) = cryoscopic constant (in \(K \cdot kg \cdot mol^{-1}\))
- \(m\) = molality of the solution (in \(mol/kg\))

### Step 1: Identify the given values
- Mass of solute (\(m_{solute}\)) = 1.00 g
- Mass of solvent (benzene) = 50 g = 0.050 kg (conversion to kg)
- Depression in freezing point (\(\Delta T_f\)) = 0.40 K
- \(K_f\) for benzene = 5.12 \(K \cdot kg \cdot mol^{-1}\)

### Step 2: Calculate the molality (m)
The molality \(m\) can be expressed as:

\[ m = \frac{n_{solute}}{mass_{solvent \, (kg)}} \]

Where \(n_{solute}\) is the number of moles of solute, calculated as:

\[ n_{solute} = \frac{mass_{solute}}{M_{solute}} \]

Where \(M_{solute}\) is the molecular mass of the solute.

### Step 3: Substitute into the depression formula
Substituting the values into the depression formula:

\[ \Delta T_f = K_f \cdot \left(\frac{n_{solute}}{mass_{solvent \, (kg)}}\right) \]

This can be rewritten as:

\[ 0.40 = 5.12 \cdot \left(\frac{\frac{1.00}{M_{solute}}}{0.050}\right) \]

### Step 4: Rearranging the equation
Rearranging the equation to solve for \(M_{solute}\):

\[ 0.40 = 5.12 \cdot \left(\frac{1.00}{M_{solute} \cdot 0.050}\right) \]

\[ 0.40 \cdot M_{solute} \cdot 0.050 = 5.12 \]

\[ M_{solute} = \frac{5.12}{0.40 \cdot 0.050} \]

### Step 5: Calculate \(M_{solute}\)
Calculating the right side:

\[ M_{solute} = \frac{5.12}{0.02} = 256 \, g/mol \]

### Final Answer
The molecular mass of the solute is **256 g/mol**.

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