2019 - JAMB Chemistry Past Questions and Answers - page 7

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.

hence, 

the oxidation numbers for Nitrogen in Ammonium Nitrite are -3, +3.

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

Soda-lime, solid is generally a white to greyish-white coloured solid. It is the mixture of calcium hydroxide and sodium or potassium hydroxide, both corrosive materials. It is noncombustible and soluble in water with release of heat. It is corrosive to metals and tissue.

The Process involves:

• Add water to the mixture and filter it. This will result in a salt solution as the filtrate; a mixture of iodine and sand as the residue.
• Evaporate the filtrate to dryness to get sodium chloride (salt).
• Heat the residue to sublime the iodine from the sand.

That is,

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

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 above 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\(_3\).

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

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

For X to liberate CO\(_2\), X must be carbonate or an oxalate. Since X decolorizes KMnO\(_4\), X must be an oxalate. 

Hence,

X is H\(_2\)C\(_2\)O\(_4\)

Recall:

V = \(\sqrt{\frac{3RT}{M}}\)

\(\therefore V \propto \frac{1}{\sqrt{M}}\)

\(V = \frac{k}{\sqrt{M}}\)

V = \(\frac{k}{M^{\frac{1}{2}}}\)

\(^{32} _{15}\)P \(\to\) 2\(^{0} _{-1}\)e + \(^{a} _{b}\)X

32 = 2(0) + a

a = 32

15 = 2(-1) + b

b = 15 + 2

b = 17

\(\therefore\) \(^{a} _{b}\)X = \(^{32} _{17}\)Cl

By applying Graham's Law of Diffusion,

\(\frac{R_1}{R_2} = \sqrt{\frac{m_2}{m_1}}\)

\(\frac{R_H}{R_O} = \sqrt{\frac{m_O}{m_H}}\)

\(\frac{R_H}{R_O} = \sqrt{\frac{32}{2}}\)

\(\frac{R_H}{R_O} = \sqrt{16} = 4\)

\(\therefore R_H = 4 \times R_O\)

Therefore, hydrogen diffuses 4 times faster than oxygen.

Graham's law of effusion was formulated by Scottish physical chemist Thomas Graham in 1848. Graham found experimentally that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles. Wikipedia