What is the normal boiling point of of a solution containing 64.5g of non volatile quinoline ( MW 129 ) in 500 g of benzene when the normal boiling point is 80.10 and KB = 2.50

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

girl imma just put my comment here because i need a couple point okurrttt