JEE MAIN - Chemistry (2018 - 15th April Morning Slot - No. 24)
The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution (Ksp of PbCl2 = 3.2 $$ \times $$ 10-8 atomic mass of Pb = 207 u ) is :
0.36 L
17.98 L
0.18 L
1.798 L
Explanation
Given,
Ksp of PbCl2 = 3.2 $$ \times $$10$$-$$8
Ksp = [s] [2s]2
$$\therefore\,\,\,\,$$ Ksp = 4s3
$$\therefore\,\,\,\,$$ 4s3 = 3.2 $$ \times $$ 10$$-$$8
$$ \Rightarrow $$ $$\,\,\,\,$$ s3 = 8 $$ \times $$ 10$$-$$9
$$ \Rightarrow $$ $$\,\,\,\,$$ s = 2 $$ \times $$ 10 $$-$$3 M
Solubility = $${{no.\,\,of\,moles\,\,of\,\,pbc{l_2}\,} \over {volume\,(in\,\,litre)}}$$
$$\therefore\,\,\,\,$$ 2 $$ \times $$ 10$$-$$3 = $${{0.1} \over {278}}$$ $$ \times $$ $${1 \over V}$$
$$ \Rightarrow $$ $$\,\,\,\,$$ V = $${{0.1} \over {278}}$$ $$ \times $$ $${{{{10}^3}} \over 2}$$ = 0.18 L
Ksp of PbCl2 = 3.2 $$ \times $$10$$-$$8
| PbCl2 $$\rightleftharpoons$$ | Pb2+ | 2Cl | |
|---|---|---|---|
| Initially | 1 | 0 | 0 |
| At equilibrium | 1-s | s | 2s |
Ksp = [s] [2s]2
$$\therefore\,\,\,\,$$ Ksp = 4s3
$$\therefore\,\,\,\,$$ 4s3 = 3.2 $$ \times $$ 10$$-$$8
$$ \Rightarrow $$ $$\,\,\,\,$$ s3 = 8 $$ \times $$ 10$$-$$9
$$ \Rightarrow $$ $$\,\,\,\,$$ s = 2 $$ \times $$ 10 $$-$$3 M
Solubility = $${{no.\,\,of\,moles\,\,of\,\,pbc{l_2}\,} \over {volume\,(in\,\,litre)}}$$
$$\therefore\,\,\,\,$$ 2 $$ \times $$ 10$$-$$3 = $${{0.1} \over {278}}$$ $$ \times $$ $${1 \over V}$$
$$ \Rightarrow $$ $$\,\,\,\,$$ V = $${{0.1} \over {278}}$$ $$ \times $$ $${{{{10}^3}} \over 2}$$ = 0.18 L
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