An aqueous solution of glucose boils at 100.01^ C . The number of glucose molecules in a solution containing 100 ~g of water is _________ [ K _ b for water is 0.5 ~K ~kg ~mol ⁻¹ ]

Step-by-step solution

The elevation in boiling point is given by T_b = T_b - T_b^ = 100.01^ C - 100^ C = 0.01 K . Using the formula T_b = K_b m , where m is the molality of the solution: 0.01 = 0.5 m m = 0.01 0.5 = 0.02 mol kg ⁻¹ . Molality m is defined as the number of moles of solute per kilogram of solvent. Given the mass of water is 100 g = 0.1 kg , the number of moles of glucose n is: n = m mass of solvent in kg = 0.02 0.1 = 0.002 moles . The number of glucose molecules is calculated by multiplying the number of moles by Avogadro's