An aqueous solution freezes at -0.36 deg C. Kf and Kb for water are 1.8 and 0.52 K kg per mole respectively then value of boiling point of solution at 1 at 1 atm pressure is

(a)101.04 deg C

(b)100.104 deg C

(c)0.104 deg C

(d)100 deg C

To determine the boiling point of the solution at 1 atm pressure, we can use the concept of colligative properties, which include freezing point depression and boiling point elevation. Since we know the freezing point depression, we can first calculate the molality of the solution and then use that to find the boiling point elevation.

Step 1: Calculate Molality from Freezing Point Depression

The formula for freezing point depression is given by:

ΔTf = Kf × m

Where:

• ΔTf is the change in freezing point (0 - (-0.36) = 0.36 °C)
• Kf is the freezing point depression constant for water (1.8 °C kg/mol)
• m is the molality of the solution

Rearranging the formula to solve for molality (m), we have:

m = ΔTf / Kf

Substituting the known values:

m = 0.36 °C / 1.8 °C kg/mol = 0.2 mol/kg

Step 2: Calculate Boiling Point Elevation

Next, we can find the boiling point elevation using the formula:

ΔTb = Kb × m

Where:

• ΔTb is the change in boiling point
• Kb is the boiling point elevation constant for water (0.52 °C kg/mol)

Substituting the values we have:

ΔTb = 0.52 °C kg/mol × 0.2 mol/kg = 0.104 °C

Step 3: Determine the New Boiling Point

The normal boiling point of pure water is 100 °C. To find the boiling point of the solution, we add the boiling point elevation to the normal boiling point:

New Boiling Point = 100 °C + ΔTb

New Boiling Point = 100 °C + 0.104 °C = 100.104 °C

Final Answer

Thus, the boiling point of the solution at 1 atm pressure is 100.104 °C, which corresponds to option (b).