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1

### 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
2

### 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.
3

### 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%
4

### 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.

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

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Class 12