1

### 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%
2

### NEET 2016 Phase 1

Consider the following liquid-vapour equilibrium.
Liquid $\rightleftharpoons$ Vapour
Which of the following relations is correct ?
A
${{d\ln P} \over {d{T^2}}} = {{ - \Delta {H_v}} \over {{T^2}}}$
B
${{d\ln P} \over {dT}} = {{\Delta {H_v}} \over {R{T^2}}}$
C
${{d\ln G} \over {d{T^2}}} = {{\Delta {H_v}} \over {R{T^2}}}$
D
${{d\ln P} \over {dT}} = {{ - \Delta {H_v}} \over {RT}}$

## Explanation

This is Clausius-Clapeyron equation.

${{d\ln P} \over {dT}} = {{\Delta {H_v}} \over {R{T^2}}}$
3

### AIPMT 2015 Cancelled Paper

If the value of equilibrium constant for a particular reaction is 1.6 $\times$ 1012, then at equilibrium the system will contain
A
mostly products
B
similar amounts of reactants and products
C
all reactants
D
mostly reactants.

## Explanation

As, 1.6 × 1012 is very high value of K. Thus, the reaction proceeds almost to completion and mixture must contain mostly products.
4

### AIPMT 2015 Cancelled Paper

The Ksp of Ag2CrO4,  AgCl,  AgBr  and Agl  are respectively, 1.1 $\times$ 10$-$12, 1.8 $\times$ 10$-$10, 5.0 $\times$ 10$-$13, 8.3 $\times$ 10$-$17. Which one of the following salts will precipitate last if AgNO3 solution is added to the solution containing equal moles of NaCl, NaBr, Nal and Na2CrO4?
A
AgBr
B
Ag2CrO4
C
Agl
D
AgCl

## Explanation

From the Ksp values of the given salts calculate the solubility values. Salt having highest solubility will precipitate at last.
AgCrO4 2Ag+ + CrO42-
s 2s s

Ksp = (2s)2(s) = 1.1 × 10–12

$\Rightarrow$ s = 0.65 × 10–4

AgCl Ag+ + Cl-
s s s

Ksp = s × s

$\Rightarrow$ 1.8 × 10–10 = s 2

$\Rightarrow$ s = 1.34 × 10–5

AgBr Ag+ + Br-
s s s

Ksp = s × s

$\Rightarrow$ 5 × 10–13 = s2

$\Rightarrow$ s = 0.71 × 10–6

AgI Ag+ + I-
s s s

Ksp = s × s

$\Rightarrow$ 8.3 × 10–17 = s2

$\Rightarrow$ s = 0.9 × 10–8

$\therefore$ Solubility of Ag2CrO4 is maximum so, it will precipitate at last.