Previous Year Question Papers NEET 2016 Phase II- Chemistry

CH₃CH₂CH₂Br + NaCN → CH₃CH₂CH₂CN + NaBr

This is a S_N2 Reaction for which polar aprotic medium is suitable for faster rate of reaction.

During hydrogenation of α, β unsaturated carbonyl compound by Pd catalyst selective reduction is observed of double bond.

α-H at bridge head carbon never show tautomerism.

Acidic strength α –I, –M effect

–I effect depend upon distance so II have stronger –I effect than III.

Bromine naturally exist in liquid phase and in question bromine is given in gaseous phase.

Reaction takes place at high temperature so mechanism with Br₂(g) should be free radical substitution which can take place in propene and butane.

Rate of free radical substitution reaction depend upon stability of free radical.

Propenyl free radical is more stable so answer will be C₃H₆

Cellulose is example of intramolecular H-bonding.

$\lambda^{o}_{M} = \dfrac{k \times 1000}{M} = \dfrac{5.76 \times 10^{-3} \times 1000}{0.5}$

= 11.52 S cm² mol^-1

Rate = k[PH3]

It is dependent on surface coverage because at low pressure, surface area covered is proportional to partial pressure of PH₃ and it is first order w.r.t PH₃.

θ = $\dfrac{kP}{1 + kP}$

θ = fraction of surface area covered at low pressure

θ = k P

$Coagulation \ power \propto \dfrac{1}{Coagulation \ value}$

Higher the coagulation power, lower is the coagulation values in millimoles per litre.

MgSO₄ > BaCl₂ > NaCl.

At Cathode:

At anode:

$\dfrac{W}{E} = \dfrac{It}{96500}$

0.1 × 2 = $\dfrac{3 × t(sec)}{96500}$

t = 6433 sec

t = 107.2 min

∼ 110 min.

3p orbital can have only 2 electrons.

$\triangle S_{sys} = nR \ ln\dfrac{P_1}{P_2} \ + \ nCP\ ln\dfrac{T_2}{T_1}$

In isothermal process T₁ = T₂

$\triangle S_{sys} = nR \ln\dfrac{P_i}{P_f}$

Ba(OH)₂ → Ba⁺² + 2OH^–

i = 3

C₅H₅N + H₅O → C₅H₅N⁺H + OH^–

$α = \sqrt{\dfrac{K_b}{c}} = \sqrt{\dfrac{1.7 \times 10^{-9}}{0.1}} = \sqrt{1.7 \times 10^{-8}}$

% α = 1.3 × 10^–4 ×100

= 1.3 × 10^–2 = 0.013.

Ca²⁺ is surrounded by 8F^–

F^– is surrounded by 4Ca⁺²

$E_{cell}^0$ < 0, so it is a non-spontaneous process

ΔG⁰ = – nFE⁰ = +ve, so ΔG⁰ > 0

ΔG⁰ = – 2.303RT log K

So, K < 1

For ideal solution inter molecular forces are identical so,

ΔH_mix = 0, ΔU_mix = 0, ΔG_mix < O

So, 1^st option is incorrect.

Ag⁺ is S

Cl^– is 0.1

K_sp = [Ag⁺] [Cl^–]

1.6 × 10^–10 = S × 0.1

1.6 × 10^–9 = S

For XY₂     n = $\dfrac{w}{M}$

$0.1 = \dfrac{10}{X + 2Y}$

X + 2Y = 100     .......(1)

For X ₃Y₂     n = $\dfrac{w}{M}$

$0.05 = \dfrac{9}{3X + 2Y}$

3X + 2Y = 180     .......(2)

Form (1) and (2)

2X = 80

X = 40 and

2Y = 100 – 40 = 60

Y = 30