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JAMB UTME - Chemistry - 2019

Question 1 Report

\(200\text{cm}^3\) of \(0.50\text{mol}/\text{dm}^3\) solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is

10g

25g

20g

15g

Answer Details

To solve this problem, we need to use the concept of stoichiometry and the solubility product constant (Ksp) of calcium hydrogen trioxocarbonate (IV). First, we need to write the balanced equation for the reaction that occurs when the solution of calcium hydrogen trioxocarbonate (IV) is heated: Ca(HCO3)2(s) → CaCO3(s) + H2O(g) + CO2(g) From the balanced equation, we can see that 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate. Therefore, we need to determine the number of moles of calcium hydrogen trioxocarbonate (IV) in the solution: Number of moles = concentration x volume Number of moles = 0.50 mol/dm³ x 0.2 dm³ Number of moles = 0.1 mol Since 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate, the number of moles of calcium carbonate produced will also be 0.1 mol. Next, we need to use the solubility product constant (Ksp) of calcium carbonate to determine the maximum amount of solid that can be precipitated: Ksp = [Ca²⁺][CO3²⁻] Ksp = 3.3 x 10⁻⁹ (at 25°C) At the maximum amount of solid precipitated, all the calcium carbonate formed will have precipitated, and the concentration of calcium ions and carbonate ions will be equal. Therefore, we can assume that the concentration of calcium ions and carbonate ions is both x. Substituting into the Ksp expression: Ksp = x² 3.3 x 10⁻⁹ = x² x = 5.74 x 10⁻⁵ mol/dm³ The mass of calcium carbonate precipitated can now be calculated: Mass = number of moles x molar mass Mass = 0.1 mol x 100.1 g/mol Mass = 10.01 g Therefore, the maximum weight of solid precipitated is approximately 10 g. Note that this calculation assumes that all the calcium carbonate precipitated as a solid, which may not always be the case in a real-world experiment. Additionally, this calculation does not take into account any losses due to filtration or other experimental errors.

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

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Question 6 Report

In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm\(^{3}\) of 1.25 molar tetraoxosulphate (vi) acid?

25cm

^{3}

10cm

^{3}

20cm

^{3}

50cm

^{3}

Answer Details

Equation of reaction : 2NaOH + H $_{2}$ SO $_{4}$ → Na $_{2}$ SO $_{4}$ + 2H $_{2}$ O

Concentration of a base, CB = 0.5M

Volume of acid, V $_{A}$ = 10cm $^{3}$

Concentration of an acid, C $_{A}$ = 1.25M

Volume of base, V $_{B}$ = ?

Recall:
$\frac{C_{A} V_{A}}{C_{B} V_{B}} = \frac{n_{A}}{n_{B}}$ ... (1)
N.B: From the equation,
$\frac{n_{A}}{n_{B}} = \frac{1}{2}$

From (1)
$\frac{1.25 \times 10}{0.5 \times V_{B}} = \frac{1}{2}$
$\frac{12.5}{0.5 V_{B}} = \frac{1}{2}$
25 = 0.5V $_{B}$

VB = 50.0 cm $^{3}$

Read lesson note on Stoichiometry And Chemical Reactions (WAEC)

Stoichiometry And Chemical Reactions

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Question 7 Report

Methane is prepared in the laboratory by heating a mixture of sodium ethanoate with soda lime. The chemical constituent(s) of soda lime is/are

Ca(OH)

_{2}

and NaOH

Na

_{2}

CO

_{3}

NaCl

NaOH

Chemical Combination

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Question 8 Report

Which of the following describes the chemical property of acids?

None of the above

Acid + XCO

_{3}

\to

Salt + H

_{2}

O + NH

_{3}

Acid + NaOH

\to

Normal salt + CO

_{2}

Acid + X

_{n}

CO

_{3}

\to

Salt + H

_{2}

O + CO

_{2}

Acids, Bases And Salts

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Question 12 Report

Which of the following conditions will most enhance the spontaneity of a reaction?

Δ

H is negative and greater than T

Δ

S

Δ

H is negative and

Δ

S = 0

Δ

H is positive and less than T

Δ

S

Δ

H is positive and equal to T

Δ

S

Kinetic Theory Of Matter And Gas Laws

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Question 15 Report

The hybridization in the compound \( \mathrm{CH_3-CH_2-C\equiv H} \) is

sp

^{3}

and sp

sp

sp

^{2}

sp

^{3}

and sp

^{2}

Answer Details

The hybridization in a and b is sp $^{3}$ hybridization while in c and d is sp hybridization.

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Question 16 Report

Which of the following is a set of neutral oxides?

N

_{2}

O, H

_{2}

O, CO

N

_{2}

O, SO

_{2}

, CO

N

_{2}

O, Na

_{2}

O, ZnO

N

_{2}

O, SO

_{3}

, NO

Oxidation And Reduction

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Question 18 Report

What mass of magnesium would be obtained by passing a current of 2 amperes for 2 hours, through molten magnesium chloride?
[1 faraday = 96500C, Mg = 24]

1.12g

4.48g

1.80g

2.00g

Answer Details

Current (I) = 2A; Time (t) = 2 hours = 7200 secs
Q = It
= 2 x 7200 = 14400C
1 F = 96500C
x = 14400C
x = $\frac{14400}{96500}$
= 0.15F
Mg $^{2 +}$ + 2e $^{-}$ $\to$ Mg
2F $\to$ 24g
0.15F $\to$ x
2x = 24 x 0.15
x = $\frac{24 \times 0.15}{2}$
= 1.8g

Read lesson note on Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Read lesson note on Electrolysis (JAMB)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Electrolysis

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Question 20 Report

When chlorine water is exposed to bright sunlight, the following products are formed

HCl

_{(a q)}

and Cl

_{2}

HOCl

_{(a q)}

and Cl

_{2}

HCl

_{(a q)}

and O

_{2}

HOCl

_{(a q)}

and O

_{2}

Chemistry Of Carbon Compounds (Ghana Only)

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Question 23 Report

X is a substance which liberates \( \mathrm{CO_2} \) on treatment with concentrated \( \mathrm{H_2SO_4} \). A warm solution of X can decolorize acidified \( \mathrm{KMnO_4} \). X is

HCOOH

NaHCO

_{3}

FeCO

_{3}

H

_{2}

C

_{2}

O

_{4}

Answer Details

It should be noted that for X to liberate CO $_{2}$ , X must be a carbonate or an oxalate. Since X decolorizes KMnO $_{4}$ , X must be an oxalate.
Therefore, X is H $_{2}$ C $_{2}$ O $_{4}$ .

Read lesson note on Periodic Chemistry (Ghana Only) (WAEC)

Periodic Chemistry (Ghana Only)

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Question 24 Report

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. If the molar mass of the compound is 180. Find the molecular formula.

[H = 1, C = 12, O = 16]

C

_{3}

H

_{6}

O

_{3}

C

_{6}

H

_{6}

O

_{3}

C

_{6}

H

_{12}

O

_{6}

CH

_{2}

O

Chemistry Of Carbon Compounds (Nigeria Only)

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Question 25 Report

For the general equation of the nature

XP + yQ ⇌ mR + nS, the expression for the equilibrium constant is

k[P]

^{x}

[Q]

^{y}

\frac{m [R] [S]}{x [P] y [Q]}

\frac{[P]^{x} [Q]^{y}}{[R]^{m} [S]^{n}}

\frac{[R]^{m} [S]^{n}}{[P]^{x} [Q]^{y}}

Answer Details

The expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is: Kc = [R]m[S]n / [P]x[Q]y where Kc is the equilibrium constant, [R] and [S] are the concentrations of the products, and [P] and [Q] are the concentrations of the reactants, all raised to the stoichiometric coefficients (m, n, x, y) in the balanced equation. This equation is known as the equilibrium constant expression and it represents the ratio of the concentrations of the products and reactants at equilibrium for a particular chemical reaction. The equilibrium constant is a measure of how far a reaction proceeds towards completion, with a larger value indicating a greater extent of reaction. The equilibrium constant expression is derived from the law of mass action, which states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants raised to their stoichiometric coefficients. At equilibrium, the rates of the forward and reverse reactions are equal, and the equilibrium constant expression represents the ratio of the rate constants for these two reactions. Therefore, the correct expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is Kc = [R]m[S]n / [P]x[Q]y.

Read lesson note on Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Read lesson note on Chemical Equilibria (JAMB)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Chemical Equilibria

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Question 26 Report

The two ions responsible for hardness in water are

Ca

^{2 +}

and / or Mg

^{2 +}

K

^{+}

and / or Mg

^{2 +}

Ca

^{2 +}

and / or Li

^{+}

Ca

^{2 +}

and / or Na

^{+}

View Answer

Question 27 Report

The cost of discharging 6.0g of a divalent metal, X from its salt is ₦12.00. What is the cost of discharging 9.0g of a trivalent metal, Y from its salt under the same condition?

[X = 63, Y = 27, 1F = 96,500C]

N12.00

N27.00

N63.00

N95.00

Answer Details

For X: X $^{2 +}$ + 2e $^{-}$ $\to$ X
2F = 63g
xF = 6g
x = $\frac{6 \times 2}{63} = \frac{4}{21} F$
$\frac{4}{21}$ F = N12.00
1F = $\frac{12}{\frac{4}{21}}$
= N63.00
1F is equivalent to N63.00.
For Y: Y $^{3 +}$ + 3e $^{-}$ $\to$ Y
3F = 27g
xF = 9g
x = $\frac{3 \times 9}{27}$
= 1F
1F = N63.00

Read lesson note on Chemical Combination (JAMB)

Chemical Combination

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Question 31 Report

Consider the equation below:

\(\mathrm{Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O}\).

The oxidation number of chromium changes from

+5 to +3

+6 to +3

-2 to +3

+7 to +3

Answer Details

Cr $_{2}$ O $_{7}^{2 -}$ + 6Fe $^{2 +}$ + 14H $^{+}$ $\to$ 2Cr $^{3 +}$ + 6Fe $^{3 +}$ + 7H $_{2}$ O
The oxidation of Cr in Cr $_{2}$ O $_{7}^{2 -}$ :
Let the oxidation of Cr = x;
2x + (-2 x 7) = -2 $⟹$ 2x - 14 = -2
2x = 12 ; x = +6
Hence, the change in oxidation of Cr = +6 to +3

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

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Question 32 Report

At 27°C, 58.5g of sodium chloride is present in 250cm\(^3\) of a solution. The solubility of sodium chloride at this temperature is?

(molar mass of sodium chloride = 111.0gmol\(^{-1}\))

2.0 moldm

^{- 3}

0.25 moldm

^{- 3}

1.0 moldm

^{- 3}

0.5 moldm

^{- 3}

Answer Details

Given the Mass of the salt = 58.5g
Volume = 250 cm $^{3}$ = 0.25 dm $^{3}$
Mass concentration = $\frac{M a s s}{V o l u m e}$
= $\frac{58.5}{0.25}$ = 234 gdm $^{- 3}$
Solubility (in moldm $^{- 3}$ = $\frac{234}{111}$
= 2.11 moldm $^{- 3}$
$≊$ 2.0 moldm $^{- 3}$

Read lesson note on Solubility Of Substances (Ghana Only) (WAEC)

Solubility Of Substances (Ghana Only)

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Question 34 Report

Which of the following could not be alkane?

C

_{4}

H

_{10}

C

_{5}

H

_{12}

C

_{7}

H

_{14}

C

_{8}

H

_{18}

Chemistry Of Carbon Compounds (Nigeria Only)

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Question 37 Report

If the cost of electricity required to discharge 10g of an ion \(X^{3+}\) is N20.00, how much would it cost to discharge 6g of ion \(Y^{2+}\)?

[1 faraday = 96,500C, atomic masses are X = 27, Y = 24]

N10.00

N6.00

N20.00

N9.00

Answer Details

X $^{3 +}$ + 3e $^{-}$ $\to$ X
3F = 27g
xF = 10g
$\frac{x}{3} = \frac{10}{27} ⟹ x = \frac{10}{9} F$
$\frac{10}{9}$ F $\equiv$ N20.00
1F is equivalent to x
$\frac{1}{\frac{10}{9}} = \frac{x}{20}$
$\frac{9}{10} = \frac{x}{20} ⟹ x = N 18.00$
1F is equivalent to N18.00.
Y $^{2 +}$ + 2e $^{-}$ $\to$ Y
2F = 24g
xF = 6g
x = $\frac{6 \times 2}{24} = \frac{1}{2} F$
1F = N18.00
$\frac{1}{2}$ F = $\frac{1}{2} \times N 18.00$
= N9.00

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Question 38 Report

By what amount must the temperature of 200cm\(^{3}\) of Nitrogen at 27°C be increased to double the pressure if the final volume is 150cm\(^{3}\) (Assume ideality)

150°C

177°C

75°C

120°C

Answer Details

Using the ideal gas law and equation:
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{P_{1} \times 200 {cm}^{3}}{300 K} = \frac{2 P \times 150 {cm}^{3}}{T_{2}}$
Cross multiply:
$T_{2} = \frac{300 \times 150 \times 2 P}{200 \times P}$
$= 450 K$ or $177^{\circ} C$
Don't forget to convert to $^{\circ} C$

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

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