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JAMB UTME - Chemistry - 2017

Question 1 Report

The oxidation number of iodine in KIO3 is

+7

+3

+5

+6

Answer Details

The oxidation number of an atom is a measure of the number of electrons that it has gained or lost in a chemical reaction. To determine the oxidation number of iodine in KIO3, we need to use the fact that the sum of the oxidation numbers of all the atoms in a compound is equal to the overall charge of the compound. In KIO3, potassium (K) has an oxidation number of +1 because it belongs to the alkali metal group and it always has a +1 oxidation state. Oxygen (O) usually has an oxidation number of -2, unless it is combined with a more electronegative atom or a peroxide, which is not the case in KIO3. Let's assume that iodine (I) has an oxidation number of x. Since KIO3 is a neutral compound, the sum of the oxidation numbers of all its atoms must be zero. Therefore, we can write the following equation: (+1) + x + 3(-2) = 0 Simplifying the equation, we get: x = +5 Therefore, the oxidation number of iodine in KIO3 is +5.

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

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Question 2 Report

The densities of two gases, X and Y are 0.5gdm-3 and 2.0gdm-3 respectively. What is the rate of diffusion of X relative to Y?

0.1

0.5

2.0

4.0

Answer Details

The rate of dimension of a gas inversely proportional to the square root of its molecular mass or its density, which is Graham's Law of diffusion of gas.

R ∝ $\frac{1}{\sqrt{M m}}$ or R ∝ $\frac{1}{\sqrt{D}}$

Dx = 0.5gdm-3, Dy = 2gdm-3

R= $\frac{K}{\sqrt{D}}$

R $\sqrt{D}$ = k

R1 $\sqrt{D_{1}}$ = R1 $\sqrt{D_{2}}$

Rx $\sqrt{D_{x}}$ = Ry $\sqrt{D_{y}}$

$\frac{R_{x}}{R_{y}}$ = $\frac{\sqrt{D_{y}}}{\sqrt{D_{x}}}$

= $\frac{\sqrt{2}}{\sqrt{0.5}}$

= 2.0

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

Kinetic Theory Of Matter And Gas Laws

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Question 3 Report

Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol-1]

0.300g

0.250g

0.2242g

0.448g

Answer Details

M =

\frac{Molar mass × Quantity of Electricity}{96500 × no of charge}

= $\frac{M m I T}{96500 n}$

Copper II Chloride = CuCl2

CuCl2 → Cu2+ + 2Cl2

Mass of compound deposited = $\frac{Molar mass × Quantity of Electricity}{96500 × no of charge}$

Q = IT

I = 0.5A

T = 45 × 60

T = 2700s

Q = 0.5 × 2700

= 1350c

Molarmass = 64gmol-1

no of charge = + 2

Mass = $\frac{64 \times 1350}{96500 \times 2}$

Mass = 0.448g

Read lesson note on Electrolysis (JAMB)

Electrolysis

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Question 13 Report

A sample of orange juice is found to have a PH of 3.80. What is the concentration of the hydroxide ion in the juice?

1.6

\times

10-4

6.3

\times

10-11

6.3

\times

10-4

1.6

\times

10-11

Answer Details

The pH scale measures the acidity or basicity of a solution. The pH is defined as the negative logarithm (base 10) of the concentration of the hydronium ion (H3O+), which is given by the expression: pH = -log[H3O+] To calculate the concentration of the hydroxide ion (OH-) in the juice, we need to use the equation for the ion product of water: Kw = [H3O+][OH-] = 1.0 × 10-14 where Kw is the ion product constant of water. At 25°C, Kw has a value of 1.0 × 10-14. To solve for [OH-], we need to rearrange the equation: [OH-] = Kw / [H3O+] Now we can substitute the pH of the orange juice to calculate the concentration of hydronium ion [H3O+]. pH = -log[H3O+] 3.80 = -log[H3O+] [H3O+] = 10^(-3.80) [H3O+] = 1.58 × 10^(-4) mol/L Substituting this value into the ion product equation: [OH-] = Kw / [H3O+] [OH-] = 1.0 × 10^-14 / (1.58 × 10^-4) [OH-] = 6.33 × 10^-11 mol/L Therefore, the concentration of the hydroxide ion in the orange juice is 6.33 × 10^-11 mol/L.

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

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Question 14 Report

Calculate the amount in moles of a gas which occupies 10.5 dm3 at 6 atm and 30oC [R = 082 atm dm3 K-1 mol-1]

2.536

1.623

4.736

0.394

Answer Details

For an ideal gas PV = nRT

Amount in moles = n

Volume v = 10.5dm3

Pressure P = 6atm

Temperature T = 30°C + 273 = 303k

R, Gas constant = 0.082 atmdm3k-1 mol

Recall from ideal gas equation

pv = nRT

n = $\frac{R V}{R T}$

n = $\frac{6 \times 105}{0.082 \times 303}$

n= 2.536mol

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

Kinetic Theory Of Matter And Gas Laws

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Question 20 Report

CH2S(g) + O2(g) → 2Cn + SO2(g)

What is the change in the oxidation number of copper in the reaction?

0 to + 2

0 to + 1

C + 1 to 0

+ 2 to + 1

Answer Details

In the reactant;

Cu2S

2 Cu - 2(1) = 0

2 Cu = 2

Cu = $\frac{2}{2}$

Cu = +1

In the product, Cu

Cu = O

The oxidation number of Cu in Cu2S and Cu respectively is +1 and 0 respectively

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

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Question 24 Report

For a general equation of the nature xP + yQ ? mR + nS, the expression for the equilibrium constant is

k [P]x [Q]y

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

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

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

Answer Details

Expression for equilibrium constant

k = $\frac{concentration of product}{concentration of reactant}$

Read lesson note on Chemical Equilibria (JAMB)

Chemical Equilibria

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Question 33 Report

X(g) + 3Y(g) ---- 2z(g) H = +ve. if the reaction above takes place at room temperature, the G will be

negative

zero

positive

indeterminate

Answer Details

$\begin{array}{cc} Enthalpy Change [ΔH] & Entropy Change [ΔS] & Gibbs free Energy[ΔG] \\ Positive & Positive & depends on T, may be + or - \\ Negative & Positive & always negative \\ Negative & Negative & depends on T, may be + or - \\ Positive & Negative & always positive \end{array}$

?G= ?H ? T?S.
To determine whether ?G will be positive or negative, the value of ?H(change in enthalpy) and ?S (change in entropy) must be given. Likewise the temperature.

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

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

Kinetic Theory Of Matter And Gas Laws

Chemical Kinetics And Equilibrium Systems (Ghana Only)

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