Question

FOR QUESTION 5-7):Solve the problems completely

5) 1.60 g of an unknown molecular substance is dissolved in 20.0 g of benzene. The freezing point of pure benzene is 5.5^oC, and the freezing point of the mixture is 2.8^oC. What is the gram formula mass of the unknown substance?

6) Complete the following two Bronsted-Lowrey reactions and indicate the acid (A), base (B), conjugate acid (CA), and conjugate base (CB). (20)

                   HSO₄^-_(aq) + H₂O_(l)   

Answer 1

5)

we know that

depression in freezing point dTf = Kf x m

so

5.5 - 2.8 = 5.12 x m

m = 0.5273

now molality = moles of solute / mass of solvent (kg)

0.5273 = moles of solute / 20 x 10-3

moles of solute = 1.054 x 10-2

now

moles = mass / molar mass

so

1.6 / molar mass = 1.054 x 10-2

molar mass = 151.70

so the molar mass of unknown substance is 151.70 g/mol

6)

HS04- + H20 ----->   H30+ + S042-

acid : HS04-

base : H20

conjugate acid : H30+

conjugate base : S042-

NH3 + H20 -----> NH4+ + OH-

acid : H20

base : NH3

conjugate base : OH-

conjugate acid : NH4+

7)

the balanced reaction is

H2S04 + 2KOH -----> K2S04 + 2H20

moles of H2S04 = mass / molar mass

moles of H2S04 = 6.1 / 98

moles of H2S04 = 6.224 x 10-2

moles of KOH = 8.2 / 56 = 1.464 x 10-1

from the above reaction

moles of KOH required = 2 x moles of H2S04

moles of KOH required = 2 x 6.224 x 10-2 = 1.244 x 10-1

but 1.464 x 10-1 moles of KOH is present

so

KOH is in excess and H2S04 is the limiting reagent

now from the above reaction

moles of water produced = 2 x moles of H2S04 reacted = 1.244 x 10-1

mass of water = moles x molar mass

mass of water = 1.244 x 10-1 x 18

mass of water produced is 2.24 g