Answer:
THE NORMAL BOILING POINT OF A SOLUTION CONTAINING 64.5 g OF QUINOLINE IN 500 g OF BENZENE IS 82.6 C
Explanation:
Boiling point = old boiling point + molarity * boiling point constant (Kb)
First, you calculate the number of moles of quinoline in the solution
number of moles = mass / molar mass
Since the molar mass of quinoline has been given as 129 g/mol
number of moles = 64.5 g / 129 g/mol
number of moles = 0.5 moles.
Next is to determine the molarity:
molarity = number of moles of solute / kilogram of solvent
molarity = 0.5 moles / 500/1000 kg
molarity = 0.5 / 0.5
molarity = 1 M
The new boiling point can then be calculated using the formula:
B. P = old B.P + Kb * molarity
B.P = 80.10 + 2.50 * 1
B.P = 80.10 + 2.50
B.P = 82.6 degree celsius
