Test Preparation on IIT JEE Mains Mock Test-6

CHEMISTRY
Given that, for the reaction H+(aq) + OH-(aq) H2O(I), energy released is 57.1 kJ. Three reactions are given as follows
(1) 0.25 mole of HCI in solution is neutralized by 0.25 mole of NaOH; heat released is ∆H1.
(2) 0.5 mole of HNO3 in solution is mixed with 0.2 mole of KOH solution; heat released is ∆H2.
(3) 200 cm3 of 0.2 M HCI solution is mixed with 300 cm3 of 0.1 M NaOH solution heat released is ∆H3.
The correct order for the numerical value of ∆H1, ∆H2, ∆H3 would be

1∆H1 >∆H2 >∆H3
2∆H1 >∆H3 >∆H2
3∆H3 >∆H2 >∆H1
4∆H2 >∆H1 >∆H3

In option (1) 0.25 mole of HCI in solution is neutralized by 0.25 mole of NaOH, heat liberated ΔH1= ¼ of 57.1 kJ. In option (2) HNO3 is in excess, neutralization reaction requires combination of 0.2 mole of HNO3 with 0.2 mole of KOH solution; heat liberated ΔH2= 1/5 of 57.1 kJ. In option (3) 200 cm3 of 0.2 M HCI contains less moles than in options 1 and 2, therefore less amount of heat is released.

Given the following standard electrode potentials:
Element Electrode Reaction
E°(298) (Volts)
Zn/Zn2+(aq) + 2e →Zn(s) - 0.76
Sn/Sn2+(aq) + 2e- →Sn(s) - 0.14
Pb/Pb2+(aq) + 2e- →Pb(s) - 0.13
Cu/Cu2+(aq) + 2e- →Cu(s) + 0.34
Based on the above data, identify the correct answer. From an aqueous solution of the salts

1zinc can displace tin, lead, copper
2tin can displace zinc, lead, copper
3copper can displace zinc, tin, lead
4lead can displace zinc, tin, copper

The standard reduction electrode potential of zinc electrode is most negative, so this metal can undergo oxidation most easily and thus can displace tin, lead and copper from their aqueous solution.

The electron energy for the quantum number n=2 in a hydrogen atom isGiven, m = 9.1 x 10-31 kg; e= 1.6 x 10-19C; h=6.6 x 10-34kgm2s-1

1- 1.312 x 106 Jmol.-1
2- 3.28 x 105 Jmol.-1
3+ 3.28 x 105 Jmol.-1
4+ 1.312 x 106 Jmol.-1

The electron energy in a hydrogen atom is given as
En =-1312/n² kj/molor -3.28 x105 J/mol

The percent void space in close fcc packing of sphere is

131.98%
225.96%
347.64%
474.04%

Volume occupied in a FCC structure = 74% the percent of void space = 100 – 74 = 26%

If the value for equilibrium constant for the reaction, A2(g) + 2 B2(g) ⇋ 4 C(g) is 2. What is the equilibrium constant for the reaction:
2 C(g) ⇋ 1/2 A2(g) + B2(g)

12
22
3√2
41/√2

For the reaction, A2(g) + 2 B2(g) ⇋ 4 C(g) k=[c]⁴/[A][B]²=For backward reaction[A][B]²/[C]⁴ for the reaction: 2 C(g) ⇋ 1/2 A2(g) + B2(g); ,K'=[A]^1/2[B]/[C]²

Solutions with reversed acidity and alkalinity are known as

1isotonic solutions
2iso-hydric solutions
3buffer solutions
4none of these

A buffer solution is one which resists changes in
pH when small quantities of an acid or an alkali
are added to it.

The rate of a second order reaction is 5 x 10-5 mol L-1s-1. When the initial concentration of the reactant is 0.25 mol L-1, the value of rate constant is

18 x 10-4 Lmol-1
22 x 10-4 molL-1
38 x 10-4 molL-1
42 x 10-4 Lmol-1

For a second order reaction; Rate = k [A]2
5 x 10-5 = k (0.25)2 = k x 0.0625
k = 5 x 10-5/0.625 x 10-1 = 8 x 10-4 Lmol-1

The reverse of chemiluminescence is called

1fluorescence
2photochemical reaction
3phosphorescence
4photosensitisation

Chemiluminescence is the emission of light
(luminescence), as the result of a chemical
reaction. Reverse of it is the absorption of light
for a chemical reaction, i.e., a photochemical
reaction.

Which of the following molecules has unpaired electron in antibonding molecular orbital?

O2 has unpaired electrons in antibonding
molecular orbital.


Right Ans	Wrong Ans	Not Attempted