Tutorial Practice Questions - Week 1

Question 1

Each statement below refers to a type of chemical bond, identify which one.

A) A hydrogen atom with a slight positive charge is attracted to a negative charge of
another molecule or atom.
B) Two atoms share electrons so they can fill their outer shells.
C) The constant motion of electrons and the creation of charge imbalances can bond
two molecules/groups together.
D) One atom loses one or more electrons and the other atom gains one or more
electrons.

Answer 1

a) H-bond

b) covalent bond

c) van der Waals interactions

d) Ionic bond

Question 2

What is the main reason for atoms to form ions?

A) To develop a positive charge
B) To fill their outer electron shells
C) To bond with other elements
D) To create a negative charge

Answer 2

B

Question 3

Which of the following has the weakest bond strength?

A) Hydrogen bonds
B) Non-polar covalent bond
C) Polar-covalent bond
D) Ionic bonds

Answer 3

A

Question 4

Balance the chemical equations

Mg + HCl –> MgCl2 + H2

Answer 4

Mg + 2HCl –> MgCl2 + H2

Question 5

Balance the chemical equations

C2H6O + O2 –> CO2 + H2O

Answer 5

C2H6O + 3O2 –> 2CO2 + 3H2O

Question 6

Balance the chemical equations

NH4NO3 –> N2O + H2O

Answer 6

NH4NO3 –> N2O + 2H2O

Question 7

Balance the chemical equations

C6H12O6 + O2 –> CO2 + H2O

Answer 7

C6H12O6 + 6O2 –> 6CO2 + 6H2O

Question 8

Which functional groups are present in the molecules below:

Answer 8

-OH, -NH3+,-COO-, -CH3

Question 9

Which functional groups are present in the molecules below:

Answer 9

3 x -OH

Question 10

Which functional groups are present in the molecules below:

Answer 10

-OPO3, -COOH

Question 11

List functional groups

Answer 11

OH, NH2

Question 12

Are these molecules soluble in water? Why?

Answer 12

Yes, they all can establish H-bonds with water.

Question 13

In the molecules below, which groups can be H-bond donors and which can be H-bonds
acceptors

Answer 13

NH3+: H donor

OH: H donor and acceptor

COO-: acceptor

Question 14

In the molecules below, which groups can be H-bond donors and which can be H-bonds
acceptors

Answer 14

P=O: acceptor

OH: donor and acceptor

C=O: acceptor

Question 15

Based on the structural formula of the following hydrocarbons, which has a double bond
in its carbon skeleton? Tip, draw the structures, and remember that carbon atoms have a
valence of 4 and should form 4 covalent bonds.

A) C3H8

B) C2H6

C) C2H4

D) C2H2

Answer 15

C) C2H4

Question 16

Which of the molecules above has the bond with the highest strength, why?

Answer 16

C2H2 because it has a triple bond between the 2 carbons.

Question 17

Calculate the molar concentration of a 415 mL solution containing 0.745 moles of HCl

𝐶=𝑛
𝑉

Answer 17

C = 0.754 mol /0.415 L
C= 1.79 M

Question 18

Calculate the molar concentration of a solution prepared by adding 34 g of NaCl to 230
mL of H2O. The molar mass of NaCl is 54.45 g/mol.

𝐶=𝑛/𝑉

𝑛=𝑚𝑎𝑠𝑠/𝑀𝑀

Answer 18

n= 34g/ 54.45g/mol = 0.62 mol

C= 0.62 mol/0.230 L = 2.7 mol/L = 2.7 M

Question 19

convert

1. 1000 mg to Kg
2. 32 g to kg
3. 0.029 kg to g
4. 40 μg to mg
5. 475 mL to L
6. 350 mL to μL
7. 3.5 mL to μL
8. 60 μL to mL
9. 5 μg/mL to mg/mL
10. 5 μg/mL to mg/L

Answer 19

1. (1g = 0.001 kg)
2. (0.032 kg)
3. (29 g)
4. (0.04 mg)
5. (0.475 L)
6. (350,000 μL)
7. (3500 μL)
8. (0.060 mL)
9. (0.005 mg/mL)
10. (0.005 mg / 0.001 L = 5 mg/L)

Question 20

How to prepare 100 mL with 0.5 mM NaCl from a stock solution of NaCl with 5 mM?

𝐶1×𝑉1 = 𝐶2×𝑉2

Answer 20

C1 = 0.5 mM
V1 = 100 mL
C2 = 5 mM
V2 = ?

V2 = C1 x V1/ C2
V2 = 0.5 mM x 100 mL/ 5 mM = 10 mL

We need 10 mL from stock solution with 5 mM NaCl nd 90 mL of water to obtain 100 mL
with 0.5 mM NaCl.

Question 21

How to prepare 150 mL with 0.2 mg/mL NaCl from a stock solution of NaCl with 50 mM?
The molar mass of NaCl is 54.45 g/mol.

Answer 21

Convert 0.2 mg/mL NaCl to molar concentration, molar is the number of moles in 1 L of
solution

C =0.2mg/mL is equal to 0.2g/L

Calculate the number of moles in 0.2 g of NaCl
n = m/M = 0.2 g / 54.45 g/mol = 0.00367 mol

C = 0.00367 mol /L = 3.67 mmol/L

C1 = 3.67 mM
V1 = 150 mL
C2 = 50 mM
V2 = C1 x V1/ C2
V2 = 3.67 mM x 150 mL/ 50 mM = 11 mL

We 11 mL from stock with 50 mM and dilute with 139 mL of water to obtain 150 mL with
3.67 mM NaCl.

Question 22

You have 4.0 mmol/L solution of albumin (molar mass = 67,000 g/mol).
Complete the following conversions of units and calculations of concentrations and
volumes. Show your answer as a number (to 1 decimal place).

(a) _____ μmoles/mL

(b) ______ g/L

(c) ____ % (weight/volume)

(d) ____ volume (in mL) containing 50 μmoles

(e) ___ mmoles in 75 mL of this solution

Answer 22

a) 4.0 mmol = 4000 μmol
1 L = 1000 mL
4000 μmol / 1000 mL = 4 μmol/mL

b) MM (albumin) = 67000 g/mol
The volume is in L, so you need to convert 4.0 mmol into mass (g).
n = mass/MM; mass = n x MM
convert mmol in mol: 4.0 = 4 x10-3 mol
mass = 4 x10-3 mol x 67000 g/ mol = 268 g
= 268 g/L

c) % weight/volume is mass (g) in 100 mL of volume
C= 268 g/L = 26.8 g/100 mL = 26.8 %
If you need more steps:
You can first determine the mass in 100 mL (100 mL = 0.1 L)
C = mass/V; replacing the numbers in this equation:
268 g/L = mass / 0.1 L; mass = 268 g/L x 0.1 L = 26.8 g
Concentration is 26.8 g in 100 mL, which is 26.8 %

d) C= n/V; V = n/C
Replacing number of moles (n = 50 μmol) and concentration (C = 4 μmol/mL)
V = 50 μmol / 4 μmol/mL = 12.5 mL

e) C= n/V; n = C x V
Replacing concentration (C = 4 μmol/mL) and Volume (V = 75 mL)
n= 4 μmol/mL x 75 mL = 300 μmol
Convert μmol to mmol (1000 μmol = 1 mmol)
300 μmol = 0.3 mmol

Question 23

You are asked to prepare a 200 mL solution of 160 mM HCl using a stock solution of
concentrated HCl with 10.25 M. What volumes of concentrated HCl and water are required
to make this solution? Show your answer in numbers (up to 2 decimal places).

(a) ____ volume of HCl stock (in mL)

b) ____ volume of water (in mL)

Answer 23

a) This is a dilution using a more concentrated solution.
C1 xV1 = C2 xV2
C1 = 160 mM HCl; V1 = 200 mL; C2 = 10.25 M, we need to find V2 and have both
concentrations in the same unit.
Converting C1 to M: 160 mM = 0.16 M
Solving the equations above for V2: V2 = C1xV1/C2;
V2 = 0.160 M x 200 mL / 10.25 M = 3.12 mL

b) volume of water to prepare 200 mL of solution
total volume = 200 mL
Volume of water = 200 mL - 3.12 mL = 196.9 mL

Question 24

3) You need to make 1 L of 10x (ten times) stock of 1 mM NaCl. (NO decimal places are
needed)

a) What would be the final concentration (mmol/L) of this 10x stock? [x1] mM

b) b) How much 10x stock NaCl would you need if you are preparing 50 mL of working solution
(1x NaCl)? [x2] mL

c) What is the dilution factor to get your stock back to working concentration? [x3]

Answer 24

a) concentration is 10 x 1 mM NaCl = 10 mM NaCl

b) C1 xV1 = C2 xV2
C1 = 1 mM; V1 = 50 mL; C2 = 10 mM, we need to calculate V2
Solving the equations above for V2: V2 = C1xV1/C2;
V2 = 1 mM x 50 mL / 10 mM = 5 mL

c) to convert 10 mM to 1 mM the stock should be diluted 10x; therefore, the dilution factor is
10.