JAMB Chemistry Past Questions & Answers

1,296.

A mixture of 0.20 mole of Ar, 0.20 mole of N2 and 0.30 mole of He exerts a total pressure of 2.1 atm. The partial pressure of He in the mixture is?

0.90 atm

0.80 atm

0.70 atm

0.60 atm

View answer & explanation

Correct answer is A

Total no mole = 0.2 + 0.2 + 0.3 = 0.7
Partial pressure of He = (0.3)/(0.7) x 2.1 =0.90

1,297.

An element, X, forms a volatile hydride XH3 with a vapour density of 17.0. The relative atomic mass of X is?

(H = 1)

34.0

31.0

20.0

14.0

View answer & explanation

Correct answer is B

Molecular Mass = Vapour density x 2 = 17.0 x 2 = 34.0
∴ Relative Atomic Mass of X in XH3 = 34 - 3
= 31.0 ∴ x = 31.0

1,298.

2.85g of an oxide of copper gave 2.52g of copper on reduction and 1.90g of another oxide gave 1.52g of copper on reduction. The data above illustrates the law of?

constant composition

conservation of mass

reciprocal proportions

multiple proportions

View answer & explanation

Correct answer is D

No explanation has been provided for this answer.

1,299.

35 cm3 of hydrogen was sparked with 12 cm3 of oxygen at 110°C and 760mm Hg to produce steam. What percentage of the total volume of gas left after the reaction is hydrogen?

11%

31%

35%

69%

View answer & explanation

Correct answer is B

Reaction for the equation:

$2H_{2} + O_{2} \to 2H_{2} O$

$2 vol : 1 vol : 2 vol$

Hence, 2 volumes of Hydrogen is needed for the reaction with Oxygen to produce steam (water).

$\therefore 12cm^{3} O_{2} \to 2 \times 12cm^{3} = 24cm^{3} H_{2}$

$\text{Excess of Hydrogen} : 35 cm^{3} - 24 cm^{3} = 11 cm^{3}$

$\text{% Hydrogen left after the reaction} = \frac{11}{35} \times 100% = \frac{220}{7} = 31.48%$

$\approxeq 31%$

1,300.

The relatively high boiling points of alkanols are due to?

ionic bonding

aromatic character

covalent bonding

hydrogen bonding

View answer & explanation

Correct answer is D

No explanation has been provided for this answer.

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