JAMB Chemistry Past Questions & Answers - Page 44

216.

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

A.

10g

B.

25g

C.

20g

D.

15g

Correct answer is A

Equation: Ca(HCO\(_3\))\(_2\) \(\to\) CaCO\(_3\) (precipitate) + H\(_2\)O + CO\(_2\) (in the presence of heat)

V = 200 cm\(^3\) = 0.2 dm\(^3\)

C = 0.5 mol/dm\(^3\) = 0.5M

N = CV \(\implies\) N = 0.5 \(\times\) 0.2

= 0.1 mole

From the equation 1 mole of Ca(HCO\(_3\))\(_2\) gives 100g of CaCO\(_3\)

(1 mole CaCO\(_3\) = 40 + 12 + (3 x 16) = 100g\)

0.1 mole of Ca(HCO\(_3\))\(_2\) gives x g of CaCO\(_3\).

\(\frac{1}{0.10} = \frac{100}{x}\)

\(x = 100 \times 0.10 = 10 g\)

217.

Which of the following sets of operation will completely separate a mixture of sodium chloride, sand and iodine?

A.

addition of water, filtration, evaporation to dryness and sublimation

B.

filtration, evaporation to dryness, addition of water and sublimation

C.

sublimation, filtration, evaporation and addition of water

D.

sublimation, addition of water, filtration and evaporation to dryness

Correct answer is A

Process:

Add water to the mixture and then filter. The resultant is a salt solution as your filtrate and a mixture of iodine and sand as your residue. Evaporate the filtrate to dryness to get your sodium chloride (salt). Heat the residue to sublime the iodine from the sand.

Addition of water \(\to\) Filtration \(\to\) Evaporation \(\to\) Sublimation.

218.

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

A.

Ca(OH)\(_2\) and NaOH

B.

Na\(_2\)CO\(_3\)

C.

NaCl

D.

NaOH

Correct answer is A

Soda lime is a mixture of caustic soda (NaOH) and lime water (Ca(OH)\(_2\)).

219.

Hydrogen bond is a sort of

A.

van der waals force

B.

dative bond

C.

ionic bond

D.

a covalent bond

Correct answer is D

Hydrogen bond is a covalent intermolecular bond that exists between hydrogen and highly electronegative elements like nitrogen, oxygen and fluorine.

220.

The oxidation state(s) of nitrogen in ammonium nitrite is/are

A.

-3, +3

B.

+5

C.

-3, +5

D.

-3

Correct answer is A

Ammonium nitrite = NH\(_4\)NO\(_2\)

NH\(_4 ^+\): Let the oxidation number of Nitrogen = x

x + 4 = 1 \(\implies\) x = 1 - 4

x = -3

NO\(_2 ^-\): x - 4 = -1

x = -1 + 4 \(\implies\) x = +3.

The oxidation numbers for Nitrogen in Ammonium Nitrite = -3, +3.