JEE Advance - Chemistry (1980 - No. 10)

(a) One litre of a sample of hard water contains 1 mg of CaCl₂ and 1 mg of MgCl₂. Find the total hardness in terms of parts of CaCO₃ per 10⁶ parts of water by weight.

(b) A sample of hard water contains 20 mg of Ca⁺⁺ ions per litre. How many milli-equivalent of Na₂CO₃ would be required to soften 1 litre of the sample?

(c) 1 gm of Mg is burnt in a closed vessel which contains 0.5gm of O₂.
      (i) Which reactant is left in excess?
      (ii) Find the weight of the excess reactants.
      (iii) How many milliliters of 0.5N H₂SO₄ will dissolve the residue in the vessel?

(a) 2 ppm; (b) 1 milli-equivalent; (c) (i) Mg; (ii) 0.25g; (iii) 166.65 ml

(a) 1 ppm; (b) 2 milli-equivalent; (c) (i) O2; (ii) 0.5g; (iii) 50 ml

(a) 0.5 ppm; (b) 0.5 milli-equivalent; (c) (i) Mg; (ii) 0.1g; (iii) 200 ml

(a) 2 ppm; (b) 1 milli-equivalent; (c) (i) Mg; (ii) 0.5g; (iii) 100 ml

(a) 1.5 ppm; (b) 0.75 milli-equivalent; (c) (i) O2; (ii) 0.125g; (iii) 75 ml

Explanation

IIT-JEE 1980 Chemistry - Some Basic Concepts of Chemistry Question 50 English Explanation

111 mg CaCl₂ will give CaCO₃ = 100 mg

1 mg CaCl₂ will give CaCO₃ = $${{100} \over {111}}$$ mg

Similarly,

1 mg MgCl₂ will give $$CaC{O_3} = {{100} \over {95}}$$ mg

Total CaCO₃ formed due to 1 mg CaCl₂ and 1 mg of MgCl₂

$$ = {{100} \over {111}} + {{100} \over {95}}$$ mg = 2 mg

CaCO₃ per litre of water = 2 mg

Weight of 1 ml of water = 1 g = 10³ mg

Weight of 1000 ml of water = 10³ $$\times$$ 10³ mg = 10⁶ mg

2 mg of CaCO₃ is present in 10⁶ mg of water.

or 2 parts of CaCO₃ in 10⁶ part of water.

(b) $$C{a^{2 + }} + N{a_2}C{O_3} \to CaC{O_3} + 2N{a^ + }$$

Equivalent weight of $$C{a^{2 + }} = {{40} \over 2} = 20$$

1 milliequivalent of Ca²⁺ = 20 mg

1 milliequi. of Ca²⁺ = 1 millieq. of Na₂CO₃

$$\therefore$$ 1 milliequivalent of Na₂CO₃ is required to soften 1 litre of hard water.

$$\mathop {2\,mg}\limits_{2 \times 24 = 48\,g} + \mathop {{O_2}}\limits_{32\,g} \to \mathop {2MgO}\limits_{2[24 + 16] = 80\,g} $$

(i) According to the above equation, 48 g Mg requires O₂ = 32g

1 g Mg requires O₂ = $${{32} \over {48}}$$ g = 0.66 g

Since the oxygen available is only 0.50 g, so whole of magnesium will not burn. Thus magnesium is in excess.

(ii) Again from the balanced equation, 32 g O₂ reacts with magnesium = 48 g

0.5 g O₂ reacts with magnesium

$$ = {{48} \over {32}} \times 0.5 = 0.75$$ g

Hence, weight of magnesium in excess

= (1.0 $$-$$ 0.75) = 0.25 g

(iii) Unused magnesium dissolves in H₂SO₄ according to the following equation:

$$\mathop {Mg}\limits_{24\,g} + \mathop {{H_2}S{O_4}}\limits_{98\,g} \to MgS{O_4} + {H_2}$$

$$\therefore$$ H₂SO₄ required to dissolve 0.25 g Mg

$$ = {{98} \over {24}} \times 0.25 = 1.021\,g$$

MgO formed dissolves in H₂SO₄ according to the following equation:

$$\mathop {MgO}\limits_{24 + 16 = 40\,g} + \mathop {{H_2}S{O_4}}\limits_{98\,g} \to MgS{O_4} + {H_2}O$$

MgO formed from 0.75

$$Mg = {{80} \over {48}} \times 0.75 = 1.25\,g$$

H₂SO₄ required to dissolve 1.25 g MgO

$$ = {{98} \over {40}} \times 1.25 = 3.062\,g$$

Total weight of H₂SO₄ required to dissolve the residue = 1.021 + 3.062 = 4.083 g

Strength of given

H₂SO₄ = Normality $$\times$$ Eq. wt.

= 0.5 $$\times$$ 49 = 24.5 g L^$$-$$1

Hence, 24.5 g H₂SO₄ is present in 1000 mL 4.083 g H₂SO₄ is present in

$$ = {{1000} \over {24.5}} \times 4.083$$ mL = 166.65 mL

JEE Advance - Chemistry (1980 - No. 10)

Explanation

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