A solution of glucose has been prepared by dissolving 1.8 g of glucose in 200 g of water. The boiling point of the solution obtained will be (Kb for water = 0.52 K kg mol–1)

To solve this problem, we need to use the formula for boiling point elevation, which is ΔTb = Kb * m, where ΔTb is the change in boiling point, Kb is the ebullioscopic constant, and m is the molality of the solution.

Step 1: Calculate the molality of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.

First, we need to convert the mass of glucose to moles. The molar mass of glucose (C6H12O6) is approximately 180 g/mol.

Number of moles = mass/molar mass = 1.8 g / 180 g/mol = 0.01 mol

Next, convert the mass of water to kilograms: 200 g = 0.2 kg

So, molality (m) = number of moles / mass of solvent in kg = 0.01 mol / 0.2 kg = 0.05 mol/kg

Step 2: Use the formula for boiling point elevation to find the change in boiling point.

ΔTb = Kb * m = 0.52 K kg mol–1 * 0.05 mol/kg = 0.026 K

Step 3: The boiling point of a solution is higher than that of the pure solvent. The boiling point of pure water is 100°C, so the boiling point of the solution is 100°C + ΔTb. However, since ΔTb is in Kelvin, we need to convert it to Celsius. The conversion from Kelvin to Celsius is straightforward because the size of the degree is the same in both scales. So, ΔTb = 0.026°C.

Therefore, the boiling point of the solution is 100°C + 0.026°C = 100.026°C.