multiple choice 1

Question 1

Which alternative is the best explanation of the action of detergents?
(A) The intermolecular bonding capabilities of a detergent molecule are due to the hydrogen atoms within the molecule.
(B) The length of a detergent molecule enables its overall polarity.
(C) The hydrophilic nature of a detergent molecule allows it to break up grease and fats.
(D) One end of the detergent molecule is polar while the other end is non-polar.

Answer 1

D - Detergents work as the non-polar section
interact with the grease/fat and the polar
part is moved through the water

Question 2

NO2(g) ——— N2O4(g)
(brown) (colourless)
NO2(g) was introduced into a closed syringe which was then capped and left for several minutes. The
plunger was then pulled out to increase the volume in the syringe.
Which observation and explanation is correct?
Observation and Explanation
(A) Colour fades Equilibrium shifts to side with more moles of gas
(B) Colour increases Equilibrium shifts to side with more moles of gas
(C) Colour increases Equilibrium shifts to side with fewer moles of gas
(D) Colour fades Equilibrium shifts to side with fewer moles of gas

Answer 2

B - Increasing volume – decrease in pressure
LCP – responds by pushing towards side with
highest number of gas moles
This is the reactants in this example so
colour would get browner (colour increases).

Question 3

What are the main intermolecular forces found in alkanes, alkenes and alkynes?

(A) Covalent
(B) Dispersion
(C) Hydrogen bonds
(D) Dipole – dipole

Answer 3

B - As these examples only contain hydrogen and
carbon (hydrocarbons) the only intermolecular
(between molecules) force is dispersion

Question 4

A 2L solution of sulphuric acid is made with 0.145g. Assuming complete dissociation, the pH of this solution would be:
(A) 2.8
(B) 3.1
(C) 0.8
(D) 3.5

Answer 4

A -  H2SO4  SO4-(aq) + 2H+(aq)
n = m/MM
MM(H2SO4) = (2 x 1.008) + 32.07 + (4 x 16.00)
                      = 98.079 g/mol
n(H2SO4) = 0.145g / 98.079 g/mol
                  =
Ratio is 2H+ for every H2SO4
n(H+) = 2 x n(H2SO4) = 
c=n/v 
c=   /2.00L
c =
pH = -log H+
pH = 2.8

Question 5

Which concentration of an HCl solution would have the same pH as a solution of 0.50 mol L-1 acetic acid? 
The Ka of acetic acid is 1.8 x 10-5.
(A) 0.090 mol L-1
(B) 0.030 mol L-1
(C) 0.0090 mol L-1
(D) 0.0030 mol L-1

Answer 5

D - CH3COOH ⇌ CH3COO- + H+
Ka = [CH3COO-] [H+]
           [CH3COOH]
Substituting x for the unknown dissociation 
concentrations
Ka =     x2     
            0.50 -x
Ignore the x in the denominator (i
(1.8 x 10-5 ) (0.50) = x2
X = 0.003
HCl will completely dissociate (strong acid) 
so [H+] will be same as x value.

Question 6

A 10.0 mL solution of NaOH with a concentration of 0.20 mol L-1 is diluted to 500 mL. What is the 
pH of the NaOH solution after dilution?
(A) 0.7
(B) 2.4
(C) 11.6
(D) 13.0

Answer 6

C - c1v1 = c2v2
0.20M x 0.0100L = c2 x 0.500L
c2 = (0.20M x 0.0100L) / 0.500L
c2 = 0.0040M
Strong base so [OH-] = 0.0040M (complete 
dissociation)
pOH = -log [OH-]
pOH = -log [0.004]
pOH = 2.40
pH + pOH = 14
pH = 14 – pOH
pH = 11.6

Question 7

What is the pH of a 0.0150 mol L-1 solution of a weak monoprotic acid that has a 
pKa = 5.5 ?
(A) 5.5
(B) 3.7
(C) 0.000218
(D) 10-5.5

Answer 7

B - pKa = -log Ka
Ka = 10-pKa
Ka = 3.26 x 10-6
HA ⇌ H+ + A-
Ka = [A-] [H+]
          [HA]
Substituting x for the unknown dissociation 
concentrations
Ka =     x2     
            0.0150 -x
Ignore the x in the denominator 
(3.26 x 10-6) (0.0150) = x2
X = 2.18 x 10-4
pH = -log[H+]
pH = 3.7

Question 8

3.50 g of ethanol was refluxed with 5.75 g of methanoic acid. The product was extracted and found
to weigh 4.23 g.
The percentage yield in this process is closest to:
(A) 34%
(B) 46%
(C) 74%
(D) 75%

Answer 8

C
n(CH3CH2OH) = 3.50g /(46.07g/mol) = 0.0760 mol
n (CHCOOH) = 5.75g / (46.025g/mol) = 0.125 mol
Limiting reagent is ethanol
Mass (ethyl methanoate) = n x MM
Mass (ethyl methanoate) = 0.0760 mol x 74.08g/mol
Mass = 5.63g
% yield = 4.23g (from question) / 5.63 (from
calculation) x100
% yield = 75%