JAMB Chemistry Past Questions & Answers - Page 49

241.

The IUPAC nomenclature of the compound
H\(_3\)C - CH(CH\(_3\)) - CH(CH\(_3\)) - CH\(_2\) - CH\(_3\)

A.

3,4- dimethylhexane

B.

2,3- dimethyl pentane

C.

2- ethyl hexane

D.

2,3- dimethylhexane

Correct answer is B

No explanation has been provided for this answer.

242.

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]

A.

C\(_3\)H\(_6\)O\(_3\)

B.

C\(_6\)H\(_6\)O\(_3\)

C.

C\(_6\)H\(_{12}\)O\(_6\)

D.

CH\(_2\)O

Correct answer is C

C → 40/12 ≈ 3

H → 6.7/1 ≈ 6

O → 53.3/16 ≈ 3

dividing through with the lowest value, 3

C\(_1\)H\(_2\)O\(_1\)

[C\(_1\)H\(_2\)O\(_1\)]n = 180

[12 + 2 + 16]m = 180

30n = 180 

n = 6

:C\(_6\)H\(_12\)O\(_6\)

243.

In the reaction:
M + N → P
ΔH = +Q kJ

Which of the following would increase the concentration of the product?

A.

Adding a suitable catalyst

B.

Increasing the concentration of P

C.

Increasing the temperature

D.

Decreasing the temperature

Correct answer is C

No explanation has been provided for this answer.

244.

Sulphur exists in six forms in the solid state. This property is known as

A.

Isomerism

B.

Allotropy

C.

Isotopy

D.

Isomorphism

Correct answer is B

No explanation has been provided for this answer.

245.

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?

A.

25cm\(^3\)

B.

10cm\(^3\)

C.

20cm\(^3\)

D.

50cm\(^3\)

Correct answer is D

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)

\(\frac{1.25 \times 10}{0.5 \times V_B} = \frac{1}{2}\)

\(\frac{12.5}{0.5V_B} = \frac{1}{2}\)

25 = 0.5V\(_B\)
VB = 50.0 cm\(^3\)