Question

. Calculate the boiling point of a solution
prepared by adding 15.00 g of NaCl to
250.0 g of water. (K, for water
= 0.512 K kg mol-?, molar mass of
NaCl = 58.44 g mol-').​

Answer 1

Answer:

100.53°C

Explanation:

As we know,

∆Tb = Tb — T°b = Km

where,

∆Tb = elevation of boiling point

Tb = boiling point of solution

T°b = Boiling point of water

Kb= elevation of boiling point constant

m = molality of solution

Given:

Mass of solute = 15g

Mass of solvent = 250g

Kb = 0.512 K kg mol-1

molar mass of solute = 58.44g

Solution:

moles of solute= n = mass / molar mass

= 15/58.44

= 0.26

Molality = (moles / mass of solvent) X 1000

= (0.26/250) X 1000

= 1.03

∆Tb = Tb — T°b = Km

Tb — 373 = 0.512 X 1.03

Tb = (0.512 X 1.03) + 373

= 0.5257 + 373

= 373.53

Therefore boiling point of solution is 373.53 K

or 100.53°C