NSMQ 2022 Quarter Final Problem of the Day Questions and Answers.

#01

A 500 cm³ solution of hydrobromic acid has a pH of 3.00. 80   cm³ of the solution was found to be completely neutralized by 20.0 cm³ of ammonia solution. Determine the concentration of the ammonia solution and the mass concentration of  the salt formed.

[N = 14 g/mol; Br = 80 g/mol; H = 1.0 g/mol]

SOLUTION:

[HBr] = 0.0010 moldm^–3

Moles of HBr = 0.0010 × 80 = 0.080 mmol

From the equation of reaction: HBr + NH₃ → NH₄Br

Mole ratio = 1:1

Hence [NH₃] = 0.080/20 = 0.0040 moldm^–3

[NH₄Br] formed =  0.080/100 = 8.0 × 10^–4

Mass concentration of NH₄Br ^­­= 8.0 × 10^–4  × 98

= 0.0784 gdm^–3  = 78.4 mgdm^–3

#02 

The reaction 2A + B → 2C was studied at 25  and the following data were obtained.

Experiment	Initial [A]/ mol dm-3	Initial [B]/ mol dm-3	Rate of formation of C/ mol dm-3 s-1
I	0.24	0.24	2.20 × 10-4
II	0.48	0.24	8.80 × 10-4
III	0.48	0.48	1.76 ×  10-3

Calculate the rate of formation of C for an experiment conducted at 25  with reactant concentrations of A = B = 0.20 mol dm^-3.

SOLUTION

Determine the rate law expression

From expt. 1 and 2, when [B] remains constant, and [A] is doubled, the rate increases four times

From expt, 2 and 3, when [A] remains constant, and [B] is doubled, the rate doubles

Hence rate law expression = Rate = K[A]²[B]

Determine the value of the rate constant

R = K[A]²[B]

Hence, K =  R/[A]²[B]. Using the values of the experiment, I (or II or III)

K= 2.2 × 10^-4 moldm^-3 s^-1/(0.24)²(0.24)mol³ dm^-9 = 1.59 × 10^-2 mol^-2 dm⁶ s^-1

Calculate the rate for the new experiment

For the experiment [A] = [B] = 0.20 mol dm^-3

Rate of formation        = 1.59 × 10^-2 mol^-2 dm⁶ s^-1 ([0.20]²[0.20]) mol³ dm^-9

= 1.27 × 10^-4 mol dm^-3 s^-1

#03

To determine the amount of ascorbic acid (vitamin C) in 500 cm³ of purchased fruit juice, a student reacts the ascorbic acid contained in the juice with an excess of iodine and back-titrates the excess iodine with a thiosulfate solution. The student reacts 25 cm³ of orange juice with 30 cm³ of 0.080 moldm^–3 of iodine. At the end of the reaction, the excess iodine required 20 cm³ of 0.20 moldm^–3 of Na₂S₂O₃. Determine the mass of ascorbic acid present in the fruit juice purchased. [C = 12 g/mol; O = 16 g/mol; H = 1.0 g/mol].

The equations of reaction are as follows:

C₆H₈O₆ + I₂  →  C₆H₆O₆  + 2H^–  + 2 I^–

I₂ + 2 S₂O₃^2–  →   2I^–   +  S₄O₆^2–

SOLUTION

Number of moles of S₂O₃^2– = 0.020 × 0.2 = 0.0040 mol

From the equation of reaction, the number of moles of I₂ = 0.0040/2 = 0.0020 mol

Moles of I₂ added = 0.080 × 0.030 = 0.0024 mol

Hence moles of I₂ which reacted = moles of C₆H₈O₆ = 0.0024 – 0.0020 = 0.00040 mol

Moles of C₆H₈O₆ in 500 cm³ of juice = 500 × 0.00040/25 = 0.0080 mol

Mass of ascorbic acid, C₆H₈O₆ = 0.0080 × 176 = 1.408 = 1.41 g

#04

The decay constant of a radioactive isotope is 1.155 × 10 ^–3 s^–1.

19. How long will it take 19.2 mg of the isotope to decay to 0.15 mg
20. Given an initial 25.6 mg sample of the radionuclide, what amount of radionuclide will decay after 3600 seconds?

[Take ln 2 as 0.693]

SOLUTION

Half-life = 0.693/1.155 × 10^–3 = 600 s = 10 minutes

19.2 mg to 0.15 mg occurs in 7 half-lives.

Hence time taken = 7 × 10 minutes = 70 minutes = 4200 seconds

3600 seconds = 60 minutes = 6 half-lives

After 6 half-lives amount of original 25.6 mg remaining will be 0.40 mg.

Amount decayed after 3600s = 25.6 – 0.40 = 25.2 mg

#05

A 0.864 g sample containing silver was dissolved in 400 cm³ of water and reacted with 30.0 cm³ of 0.500 moldm^–3 of potassium iodide solution. The resulting precipitate was carefully washed, dried to a constant mass, and weighed to obtain a mass of 0.470 g. Determine the percentage composition by mass of silver in the sample.

Given that 200 mg of silver is required to make a coin, how many coins can be made from the sample given?

[K = 39.0 g/mol; I = 127 g/mol; Ag = 108 g/mol]

SOLUTION

Precipitate = AgI

Equation of reaction = Ag⁺ + I^– → AgI_(s)

Mass of AgI = 0.470 g

Molar mass of AgI = 235 g/mol

Moles of AgI = 0.470/235 = 0.00200 mol

Mass of Ag = 0.00200 × 108 = 0.216 g

Percentage of silver in sample = 0.216/0.860 × 100 = 25.0%

The sample contains 216 mg of silver. Hence, only one coin can be made from the sample.

#06

A hydrocarbon has the molecular formula C₅H₁₀. Give the names of all possible structures of the hydrocarbon.

SOLUTION

• 1-pentene
• 2-pentene
• 2-methyl-1-butene
• 2-methyl-2-butene
• 3-methyl-1-butene
• Cyclopentane
• Methyl cyclobutane
• 1,1-dimethylcylopropane
• 1,2-dimethylcylopropane

#07

A hydrocarbon Q is made up of 85.7% carbon and 14.3% hydrogen. At 127 °C and 80.0kPa, 0.510dm³ of the hydrocarbon burns in excess oxygen to give 2.04dm³ of CO₂ and 2.04dm³ of water vapour. Calculate the molecular formula of the hydrocarbon from the data given. The Ideal Gas constant is 8.31JK^-1mol^-1.

C         =          12.0;    H         =          1.00

Ans:

C                                             H

85.7/12.0                                 14.3/1.00

7.14                                         14.3

1                                              2

Hence empirical formula        =          CH₂;

Assume Molecular formula    =          n(CH₂)            where n is an integer

Q_(g)       →        nCO_2(g)            +          nH₂O_(g)

Volumes in the combustion:   0.510 dm³ Q;   2.01dm³ CO_2(g);          2.01dm³ H₂O_(g).

Ratio of volumes in the combustion               1:4:4

Applying Gay-Lussac’s Law, the stoichiometric ratios in the combustion should be           1:4:4

Hence 1mol Q burns to give 4mol CO_2(g) and 4mol H₂O_(g)

Hence, the Molecular formula of Q                is C₄H₈.

[Note: Full marks not to be awarded if the deduction is not fully shown.]

#08

The balanced equation for the gas-phase decomposition of compound A to B and C is as follows:

2A(g)               ⇌                     2B(g)               +          C(g).

The equilibrium constant K_c of the reaction at 250 °C is 1.08*10^–11.

1. Write down the expression for the equilibrium constant K_c of the reaction.

2. Calculate the equilibrium concentration of A_(g), B_(g) and C_(g) of the reaction if the initial concentration of A_(g) is 0.100moldm^-3.

3. State a reasonable assumption made in the calculation.

4. Predict the effect of an increase in the total pressure on the equilibrium process.

Ans:

1. a) K_c =          [B]²[C]/[A]²
2. b) 2A(g) ⇌                     2B(g)               +          C(g).

Initial conc.     0.100                                       0.00                             0.00

At equil.          (0.100 – 2x)                             2x                                x

Kc       =          (2x)²(x)/(0.100 – 2x)²   =        4x³/(0.100 – 2x)².

≈          4x³/(0.100)²                 =          1.08*10^-11

4x³       =          1.08*10^-13;                  x³         =          27.0*10^-15

x          =          3.00*10^-5 moldm^-3.

Hence conc of A         =          0.100  or 1.00*10^-1     [OR     0.09994 or 9.99*10^-2] moldm^-3.

Conc of B        =          6.00*10^-5 moldm^-3;     Conc of C       =          3.00*10^-5 moldm^-3.

c) The assumption: The amount of A that decomposes at equilibrium is small and negligible compared with the initial concentration.

d) An increase in pressure will favour the reverse reaction, OR less of A will be formed at higher pressure, but the equilibrium constant K_c will not

#09

You are provided with about 10.0 g of neutral, solid organic compound X. The compound is contaminated with an organic acid. You are also informed that X is insoluble in cold ethanol but moderately soluble in hot ethanol. It is very soluble in cold ethyl ethanoate . It is insoluble in cold water but sparingly soluble in hot water.

Describe in detail how you would remove the organic acid compound from X and get X ready for recrystallisation.

State the solvent that would be most suitable for the recrystallisation of X.

Ans:

a) 1) The solid is transferred into a 200 – 400 cm³

About 50 to 100 cm³ of ethyl ethanoate is added to the solid X to dissolve it.

3)   The solution of X is transferred into a separating funnel and about 50 cm³ of dilute NaOH solution is added. Two layers are observed.

4)   The funnel is stoppered and shaken intermittently for about two minutes.

• The funnel is clamped long enough for the two layers to separate.
• The stopper is removed, the tap of the funnel is opened, and the lower (aqueous) layer is run down.
• The extraction is repeated a few (up to three) more times.
• The ethyl ethanoate solution of X is washed twice with deionized water using the separating funnel.
• The washing is repeated until the water tests neutral to litmus.
• The solution of X is transferred into a conical flask and about 5g of anhydrous Na₂SO₄ or MgSO₄ is added to dry the solution.
• After about 30 minutes, the solution is filtered into a beaker and heated/boiled to remove the solvent(The solvent may be removed by distillation)

b). Solvent for recrystallization is ETHANOL

#10

An entrepreneur in Ghana decides to set up a factory by using the large limestone deposits close to her hometown to manufacture an aqueous suspension of slaked lime, which can be used as white paint. The limestone will be subjected to three processes to obtain the intended product as follows:

1. The limestone is heated in a kiln to give lime or calcium oxide and CO₂ gas.
2. The lime is treated with a measured amount of water to convert it into solid slaked lime.
3. The slaked lime is then suspended or partially dissolved in excess water to obtain the paint.

Consider the following enthalpies of formation all in kJmol^-1and deduce if the heat to be generated from the three steps can be used to operate the factory on a sustainable basis:   CaCO_3(s)      -1206; CaO_(s)   -635;            Ca(OH)_2(s)   -986;  Ca²⁺_(aq)   -543;       CO_2(g)   -394;   H₂O_(l)   -286; OH^–_(aq)   -230.

Ans:

Step 1: CaCO_3(s)        →        CaO_(s)             +          CO_2(g).

ΔH₁(reaction)  =          (-635 + -394)   –   (-1206)       =          +177kJ.

Step 2: CaO_(s)            +          H₂O_(l)              →                    Ca(OH)_2(s)

ΔH₂(reaction)  =          -986  –  (-635  +  -286)            =          -65.0kJ.

Step 3: Ca(OH)_2(s)     +   excess water          →        Ca²⁺_(aq)             +2OH^–_(aq).

ΔH₃(reaction)  =          [-543  +2(-230)]  –  [-986]       =          -17.0kJ

Sum of the three enthalpies of reaction = +177 -65.0  – 17.0 = +95.0kJ

Hence, the three reactions will not generate any heat to sustain the three processes. Rather, external heat will be needed to keep the factory running.

[Note: If the conclusion is not based on the ΔH’s of the three processes, no marks should be awarded for the conclusion.]

#11

2,2,4-Trimethylpentane burns smoothly in internal combustion engines and, for this reason, is assigned an octane rating of 100. Its enthalpy of combustion can be calculated by;

1. i) Hess’s Law and
2. ii) the use of bond energies.

Determine the enthalpy of combustion by the two methods and comment on the relative values obtained. You may use the following enthalpies of formation and bond energies, all in kJmol^-1:

2,2,4-Trimethylpentane -225; CO_2(g)  -394;    H₂O_(g) -242;    C=O 745; C-C 350; C-H 415; O-H 464; O₂ 498.

Ans:

2,2,4-Trimethylpentane           =          C₈H₁₈.

C₈H₁₈              +          25/2O₂             →        8CO_2(g)            +          9H₂O_(g)

Enthalpy of combustion by Hess’s Law:

ΔH(combustion) =   (-394*8 + -242*9) – (-225)  =   -5105 or  -5.11*10³kJmol^-1.

Enthalpy of combustion using bond energies:

Bonds broken              =          7C-C   +          18C-H +          12.5O2

Energy required          =          350*7 + 415*18 + 498*12.5               =          16145kJ

Bonds formed             =          16C=O            +          18O-H

Energy given out        =          -745*16           +          -464*18           =          -20,272kJ

ΔHʹ(combustion)         =          16,145 – 20,272           =          -4,127  or  -4.13*10³kJmol^-1.

Enthalpy of combustion by Hess’s Law is larger than the Enthalpy of combustion using bond energies.

The former is more accurate because it is obtained using enthalpies of formation, which are experimental values, whereas bond energies are approximate (or theoretical) values.

#012

A metal Q forms an oxide when 10.4g of it reacts with 7.48dm³ of oxygen gas at 27.0^oC and 100kPa.

Determine the formula of the oxide and the percent oxygen by mass in the oxide. Atomic mass of Q is 52.0. The Ideal Gas constant is 8.31JK^-1mol^-1.

Ans:

Moles of Q made to react with oxygen = 10.4/52.0   = 0.200.

Moles of oxygen gas, n reacting, = PV/RT;   P = 100kPa;    V = 7.48dm³;  T = 300K.

= 100 X 7.48/8.31 X 300 = 0.300

Moles of oxygen atoms = 0.600

Hence formula   = QO₃

%O by mass in QO₃    = [48.0/(48.0+52)] x 100

= 48.0

#13

Gaseous compounds A and B react to give product compound C according to the following equation:

A(g)     +          B(g)                 →                    2C(g)

1. Use the following kinetic data to determine the Rate Law for the reaction

Experiment	Initial concentration in moldm-3		Initial Rate (moldm-3s-1)
	A	B
1	0.100	0.100	2.40 x 10-2
2	0.200	0.100	9.60 x 10-2
3	0.300	0.100	2.16 x 10-1
4	0.300	0.200	4.32 x 10-1
5	0.300	0.300	6.48x 10-1

1. Use the results to determine the rate constant k, for the reaction and give the units.
2. Predict the initial rate of the reaction if the initial concentrations of A and B were 0.400 and 0.500 moldm^-3, respectively.

Ans:

1) Rate Law:   Consider the results of Expts. 1 and 2;

When the concentration of A is doubled while that of B remains constant, the initial rate increases by a factor of 4.

Hence, the reaction is 2^nd order concerning A. Consider Expts. 3 and 4. It can be deduced that the reaction is 1^st order concerning B.

Hence, the Rate Law:             Rate     =   k[A]²[B].

2) Use any set of experimental results to calculate k. e.g.

Experiment 1

2.40 x 10^-2 =    k(1.00 x 10^-1)²(1.00 x 10^-1)     = k(1.00 x 10^-3)

k          = 2.40 x 10¹ dm⁶mol^-2s^-1

3) Using Expt. 1, when the concentration of A is increased from 0.100 to 0.400 moldm^-3, the rate increases by a factor of 16.

When the concentration of B is increased from 0.100 to 0.500 moldm^-3, the rate increases by a factor of 5.

Hence the total expected increase       = 5 x 16           =   80

The expected rate        = 80.0*2.40*10^-2        = 192*10^-2      = 1.92 moldm^-3s^-1

[Note:  Accept any mathematical method giving the right answer]

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