What is the amount of urea to be dissolved in 500 g of water to produce a depression of 0.186^oC in the freezing point ? (K_f for water = 1.86 K/m)

3 g

The correct answer is option 3. 3 g.

To determine the amount of urea to be dissolved in water to produce a specific depression 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 the freezing point (0.186°C in this case).

\(K_f\) is the cryoscopic constant (1.86 K·kg/mol for water).

\(m\) is the molality of the solution.

First, solve for molality \(m\):

\(m = \frac{\Delta T_f}{K_f} = \frac{0.186}{1.86} = 0.1 \, \text{mol/kg}\)

Next, determine the number of moles of urea needed to achieve this molality in 500 g (0.5 kg) of water:

\(m = \frac{\text{moles of solute}}{\text{kg of solvent}}\)

\(0.1 \, \text{mol/kg} = \frac{\text{moles of urea}}{0.5 \, \text{kg}}\)

\(\text{moles of urea} = 0.1 \, \text{mol/kg} \times 0.5 \, \text{kg} = 0.05 \, \text{mol}\)

Now, calculate the mass of urea required. The molar mass of urea (\(\text{CH}_4\text{N}_2\text{O}\)) is approximately 60 g

mol:

\(\text{mass of urea} = \text{moles of urea} \times \text{molar mass of urea} \)

\(\text{mass of urea} = 0.05 \, \text{mol} \times 60 \, \text{g/mol} = 3 \, \text{g}\)

Therefore, the amount of urea to be dissolved in 500 g of water to produce a depression of 0.186°C in the freezing point is: 3 g.