The amount of urea to be dissolved in `500` cc of water `(K_(f)=1.86)` to produce a depresssion of `0.186^(@)C` in the freezing point is :

Options

A. 0.3 gm
B. 3 gm
C. 6 gm
D. 9 gm

Answer

To solve the problem of how much urea needs to be dissolved in 500 cc of water to produce a depression of 0.186°C in the freezing point, we can use the formula for freezing point depression:

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

Where:
- \(\Delta T_f\) is the depression in freezing point (in °C).
- \(K_f\) is the cryoscopic constant (in °C kg/mol).
- \(m\) is the molality of the solution (in mol/kg).

### Step 1: Calculate the molality (m)

First, we need to express molality in terms of the mass of solute (urea) and the mass of solvent (water). The formula for molality is:

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

Where:
- \(n_{solute}\) is the number of moles of solute (urea).
- \(mass_{solvent}\) is the mass of the solvent (water) in kg.

Since we have 500 cc of water, we can convert this to kg (assuming the density of water is approximately 1 g/cc):

\[ mass_{solvent} = 500 \, \text{g} = 0.5 \, \text{kg} \]

### Step 2: Rearranging the formula

We can rearrange the freezing point depression formula to find the number of moles of urea:

\[ \Delta T_f = K_f \cdot m \implies m = \frac{\Delta T_f}{K_f} \]

Substituting the values:

\[ m = \frac{0.186}{1.86} \approx 0.1 \, \text{mol/kg} \]

### Step 3: Calculate the number of moles of urea

Now, we can find the number of moles of urea using the molality:

\[ n_{solute} = m \cdot mass_{solvent \, (kg)} = 0.1 \cdot 0.5 = 0.05 \, \text{mol} \]

### Step 4: Calculate the mass of urea

To find the mass of urea required, we use the molar mass of urea, which is approximately 60 g/mol:

\[ mass_{urea} = n_{solute} \cdot M_{urea} = 0.05 \, \text{mol} \cdot 60 \, \text{g/mol} = 3 \, \text{g} \]

### Conclusion

The amount of urea to be dissolved in 500 cc of water to produce a depression of 0.186°C in the freezing point is **3 grams**.