Test Preparation on IIT JEE Mains Mock Test-8

Amount of A (t ½ =20 min.) left after one hour or 3 half life periods = 1→1/2 →¼ → 1/8
Amount of A (t ½ =10 min.) left after one hour or 6 half life periods =
1→1/2 → ¼ → 1/8→1/16 →1/32→ 1/64

NaOH HCl
M1V1 – M2V2 = M3V3
0.45 x 10 – 0.1 x 40 = M3 x 50
4.5 – 4.0 = 0.5 = M3 x 50
M3 = 0.5/50 = 0.01 or 1 x 10-2
[OH-] = 10-2 ; [H+] = 10-12 and
pH = -log10-12 = 12

SO2 → S; change in oxidation number of
S (+4 → 0) = 4
Equivalent mass of SO2 =
Molecular mass/4 = 64/4 = 16

N2O4 ⥪ 2NO2
1 0
1 – 0.2 2 x 0.2;
Total moles = 0.8 + 0.4 = 1.2
When temperature becomes double at constant
volume, pressure will also become double, i.e.,
2 atm. since number of moles are also changing,
total pressure = 1.2 x 2 = 2.4 atm.

One Faraday will liberate 0.5 g mol of Be from
Be2+ ; 0.5 g mol of Cu from Cu2+ ; 0.33 g mol of Al
from Al3+ and 1 g mol of Na from Na+ .

On passing H2S in dilute HCl solution, cations of
group II (Hg2+, Pb2+, Cu2+, Cd2+) are precipitated
so cations from options (2), (3) and (4) can be
separated. Al3+ (group III cation) and Sn2+ are not
precipitated.

Element A can lose one of its valence electron
easily and element B can accept one electron
easily to achieve stable noble gas electronic
onfigurations. So the correct formula of the
compound formed between A and B is A+B-.

Total number of electrons in NO3
- = 7+ 24 + 1 =
32; CO3
2- = 6+ 24 + 2 = 32; ClO3
- = 17+ 24 + 1 =
42 and in SO3 = 16+ 24 = 40. Therefore, ions
NO3
- and CO3
2- are isoelectronic; also these two
have same structures, i.e., are isostructural.