-0 .3 degree Celsius... therefore option is less than 0.C option(b)The freezing point depression can be calculated by a formulaΔTf​=iKf​bΔTf​=Tfo​−Tf= the depression of freezing point.i= the van't hoff factor =3Kf​= Molar freezing depression constant = 1.86kg/molb= molar concentration of the solution = 0.062 mol/kgδTf​=3×1.86×0.062=0.34oCTf​=Tfo​−0.34o=−0.34o

Solution:

Calculation of Freezing Point Depression:

The freezing point depression of a solution is given by the formula:

ΔTf = Kf × molality

Where ΔTf is the freezing point depression, Kf is the cryoscopic constant, and molality is the molal concentration of the solution.

Cryoscopic Constant:

The cryoscopic constant for water is 1.86 °C/m.

Molality:

Molality is defined as the number of moles of solute per kilogram of solvent.

molality = (moles of solute) / (mass of solvent in kg)

Calculation of Molality:

The molecular weight of Ca(NO3)2 is 164.1 g/mol.

1% solution of Ca(NO3)2 means 1 g of Ca(NO3)2 is dissolved in 100 mL of water.

Therefore, the number of moles of Ca(NO3)2 in 1 L of solution = (1 g / 164.1 g/mol) × (1000 mL / 100 mL) = 0.0061 mol/L

The mass of water in 1 L of solution = 1000 g - 1 g = 999 g

molality = (0.0061 mol/L) / (999 g / 1000 g) = 0.0061 mol/kg

Calculation of Freezing Point Depression:

ΔTf = Kf × molality = 1.86 °C/m × 0.0061 mol/kg = 0.011346 °C

Explanation:

The freezing point depression of a solution is the difference between the freezing point of the solvent and the freezing point of the solution. Since the freezing point of pure water is 0°C, the freezing point of the solution would be less than 0°C. Therefore, the correct answer to the question is option (b), less than 0°C.

Conclusion:

The aqueous solution of Ca(NO3)2 with a concentration of 1% has a freezing point depression of 0.011346°C, which means that its freezing point is less than 0°C.