Elevation of boiling point of 1 molar aqueous glucose solution (density = 1.2 g/ml) is

To calculate the elevation of boiling point of a 1 molar aqueous glucose solution, we need to use the formula for boiling point elevation:

ΔTb = Kb * m

Where: ΔTb is the elevation of boiling point, Kb is the molal boiling point elevation constant, and m is the molality of the solution.

First, we need to calculate the molality (m) of the solution. Molality is defined as the number of moles of solute (glucose) per kilogram of solvent (water). Since we have a 1 molar solution, the molality will be 1 mol/kg.

Next, we need to find the molal boiling point elevation constant (Kb) for water. The Kb value for water is approximately 0.512 °C/m.

Now, we can substitute the values into the formula:

ΔTb = 0.512 °C/m * 1 mol/kg

Simplifying the equation, we find that the elevation of boiling point (ΔTb) for a 1 molar aqueous glucose solution is 0.512 °C.