1
MCQ (Single Correct Answer)

NEET 2016 Phase 1

MY and NY3, two nearly insoluble salts, have the same Ksp values of 6.2 $$ \times $$ 10$$-$$13 at room temperature. Which statement would be true in regard to MY and NY3?
A
The salts MY and NY3 are more soluble in 0.5 M KY than in pure water.
B
The addition of the salt of KY to solution of MY and NY3 will have no effect on their solubilities.
C
The molar solubilities of MY and NY3 in water are identical.
D
The molar solubility of MY in water is less than that of NY3.

Explanation

MY ⇌ M+ + Y
   s        s       s

Ksp = s.s = s2

$$ \Rightarrow $$ s = $$\sqrt {6.2 \times {{10}^{ - 13}}} $$

= 7.87 × 10–7 mol L–1

NY3 ⇌ N+ + 3Y
   s        s       3s

Ksp = s.(3s)3 = 27s4

6.2 × 10–13 = 27s4

$$ \Rightarrow $$ s = 3.89 × 10–4 mol L–1

Hence, molar solubility of MY in water is less than that of NY3.
2
MCQ (Single Correct Answer)

NEET 2016 Phase 2

The solubility of AgCl(s) with solubility product 1.6 $$ \times $$ 10$$-$$10 in 0.1 M NaCl solution would be
A
1.26 $$ \times $$ 10$$-$$5 M
B
1.6 $$ \times $$ 10$$-$$9 M
C
1.6 $$ \times $$ 10$$-$$11 M
D
zero

Explanation

AgCl ⇌ Ag+ + Cl
   s           s     s + 0.1

Concentration of Cl is (s + 0.1) mol L–1 because s mol L–1 from ionization of AgCl and 0.1 mol L–1 from ionization of 0.1 M NaCl.

Now, Ksp = [Ag+][Cl ]

$$ \Rightarrow $$ 1.6 × 10–10 = s (s + 0.1)

$$ \Rightarrow $$ 1.6 × 10–10 = s (0.1) {$$ \because $$ s << 0.1}

$$ \Rightarrow $$ s = 1.6 × 10–9 M
3
MCQ (Single Correct Answer)

NEET 2016 Phase 2

Which of the following fluro-compounds is most likely to behave as a Lewis base ?
A
BF3
B
PF3
C
CF4
D
SiF4

Explanation

PF3 is lewis base because on P atom there is lone pair.
4
MCQ (Single Correct Answer)

NEET 2016 Phase 2

The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5N+H) ina 0.10 M aqueous pyridine solution (Kb for C5H5N = 1.7 $$ \times $$ 10$$-$$9) is
A
0.0060%
B
0.013%
C
0.77%
D
1.6%

Explanation

C5H5N + H2O → C5H5N+H + OH

So, the amount of pyridine that forms pyridinium ion is $$\alpha $$.

Now, $$\alpha $$ = $$\sqrt {{{{K_b}} \over {conc.\,of\,pyridine}}} $$

= $$\sqrt {{{1.7 \times {{10}^9}} \over {0.10}}} $$

= 1.30 $$ \times $$ 10-4

So, percentage of pyridine that forms pyridinium ion

= 1.30 × 10–4 × 100

= 1.30 × 10–2 = 0.013%

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