The amount of benzene that wil separate out (in grams) if a solution containing `7.32 g` of triphenylmethane in `1000g` of benzene is cooled to a temperature which is `0.2^(@)C` below the freezing point of benzene?
`(K_(f)=5.12K-Kg//"mol")`

Answer

To solve the problem, we need to determine how much benzene will separate out when a solution containing 7.32 g of triphenylmethane in 1000 g of benzene is cooled to a temperature that is 0.2°C below the freezing point of benzene.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of triphenylmethane (solute) = 7.32 g
- Mass of benzene (solvent) = 1000 g
- Depression in freezing point (ΔTf) = 0.2°C
- Freezing point depression constant (Kf) for benzene = 5.12 K·kg/mol

2. **Calculate the Number of Moles of Triphenylmethane:**
- First, we need to find the molar mass of triphenylmethane (C19H16). The molar mass is calculated as follows:
- Carbon (C) = 12.01 g/mol, Hydrogen (H) = 1.008 g/mol
- Molar mass of triphenylmethane = (19 × 12.01) + (16 × 1.008) = 244.24 g/mol (approximately 244 g/mol)
- Now, calculate the number of moles of triphenylmethane:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{7.32 \text{ g}}{244 \text{ g/mol}} \approx 0.0300 \text{ mol} \]

3. **Calculate the Molality of the Solution:**
- Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0300 \text{ mol}}{1 \text{ kg}} = 0.0300 \text{ mol/kg} \]

4. **Use the Freezing Point Depression Formula:**
- The formula for freezing point depression is given by:
\[ \Delta T_f = K_f \times m \]
- Substitute the known values into the equation:
\[ 0.2 = 5.12 \times m \]
- Solve for molality (m):
\[ m = \frac{0.2}{5.12} \approx 0.0391 \text{ mol/kg} \]

5. **Calculate the Weight of Solvent (Benzene) at New Conditions:**
- Rearranging the molality formula gives us:
\[ \text{Number of moles of solute} = m \times \text{mass of solvent in kg} \]
- We can express the mass of solvent (W) in grams:
\[ 0.0300 = 0.0391 \times \frac{W}{1000} \]
- Solving for W:
\[ W = \frac{0.0300 \times 1000}{0.0391} \approx 768 \text{ g} \]

6. **Calculate the Amount of Benzene that Separates Out:**
- The initial mass of benzene was 1000 g. The mass of benzene remaining after cooling is 768 g.
- Therefore, the amount of benzene that separates out is:
\[ \text{Amount of benzene that separates} = 1000 \text{ g} - 768 \text{ g} = 232 \text{ g} \]

### Final Answer:
The amount of benzene that will separate out is **232 grams**.