°C is 96.535 mmHg. The vapor pressure of pure water at the same temperature is 760 mmHg. Calculate the vapor pressure of the solution and the boiling point elevation constant (Kb) for water.

To solve this problem, we need to use the equation:

ΔTb = Kb * molality

where ΔTb is the boiling point elevation, Kb is the boiling point elevation constant, and molality is the molal concentration of the solution.

First, we need to calculate the molality of the solution:

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

We are given the mass of glucose (18 g) and the mass of water (178.2 g). To convert the mass of water to kg, we divide by 1000:

mass of water = 178.2 g / 1000 = 0.1782 kg

Next, we need to calculate the number of moles of glucose:

moles of glucose = mass of glucose / molar mass of glucose

The molar mass of glucose (C6H12O6) is:

6 * 12.01 g/mol + 12 * 1.01 g/mol + 6 * 16.00 g/mol = 180.18 g/mol

So the number of moles of glucose is:

moles of glucose = 18 g / 180.18 g/mol = 0.0999 mol

Now we can calculate the molality:

molality = 0.0999 mol / 0.1782 kg = 0.560 mol/kg

Next, we can use the following equation to calculate the vapor pressure of the solution:

P = Xsolvent * Psolvent

where P is the vapor pressure of the solution, Xsolvent is the mole fraction of the solvent (water), and Psolvent is the vapor pressure of pure water at the same temperature.

We can assume that the glucose does not contribute significantly to the vapor pressure, so Xsolvent is approximately equal to 1 (i.e. water is the only solvent). Therefore:

P = Psolvent = 96.535 mmHg

Finally, we can use the boiling point elevation equation to calculate Kb:

ΔTb = Kb * molality

We are given that the boiling point elevation (ΔTb) is equal to the difference between the boiling point of the solution and the boiling point of pure water, which is 100°C. Therefore:

ΔTb = 100°C - 100°C = 0°C

Substituting into the equation, we get:

0°C = Kb * 0.560 mol/kg

Kb = 0°C / 0.560 mol/kg = 0°C/m

Therefore, the boiling point elevation constant (Kb) for water is 0°C/m.