1) A second order reaction is always a multistep reaction
2) A first order reaction is always a single step reaction
3) A zero order reaction is a multistep reaction
4) A zero order reaction is a single step reaction
Factual
Solution: (3)
1) 100 mL of 0.1 M HCl and 200 mL of 0.1 M CH3COONa
2) 100 mL of 0.1 M HCl and 200 mL of 0.1 M NaCl
3) 100 mL of 0.1 M CH3 COOH and 100 mL of 0.1 M NaOH
4) 100 mL of 0.1 M CH3COOH and 200 mL of 0.1 M NaOH
Solution: (1)
1) (a), (c) and (d)
2) (b) and (c)
3) (c) and (d)
4) (a) and (d)
Diazo compounds and inorganic nitrogen can’t be estimated by the kjeldahl method.
Solution: (3)
1) greater than 300 K but less than 373 K
2) equal to 373 K
3) more than 373 K
4) less than 300 K
Less than 300 K (factual)
Solution: (4)
1) cis– [CrCl2 (ox)2]3- (ox – oxalate)
2) trans – [Fe (NH3)2 (CN)4]–
3) trans – [Cr (Cl2) (ox)2]3-
4) cis– [Fe (NH3)2 (CN)4]–
cis– [CrCl2 (ox)2]3- (ox – oxalate)
Solution: (1)
has most acidic hydrogen among given compounds , this is due to the strong – M effect of –CN group which stabilize –ve charge significantly.
Solution: (3)
1) N2O3
2) N2
3) N2O5
4) NO
Solution: (4)
1) Br2 / water; ZnCl2 / HCl; FeCl3
2) Br2 / water; ZnCl2 / HCl; NaOCl
3) Alcoholic HCN; NaOCl; ZnCl2 / HCl
4) ZnCl2 / HCl; FeCl3; Alcoholic HCN
Novestrol
It can react with Br2 / water due to presence of unsaturation, with ZnCl2 / HCl due to – OH group and with FeCl3 due to phenol.
Solution: (1)
Solution: (3)
1) NO–
2) NO+
3) NO2+
4) NO
B.O. NO– = 2
BO NO+ = 3
BO NO2+ = 2.5
BO NO = 2.5
B.O ∝ 1 / [B.L]
Solution: (1)
1) Fractional distillation
2) Distillation under reduced pressure
3) Differential extraction
4) Steam distillation
conceptual
Glycerol is separated in soap industries by distillation under reduced pressure.
Solution: (2)
1) Ozone layer depletion
2) Blue baby syndrome
3) Eutrophication
4) Acid rain
Refer environmental chemistry
It emits CO2 that combines with moisture of atmosphere and forms H2CO3 (carbonic acid).
Solution: (4)
|
ɑ |
β |
𝛾 |
𝝳 |
|
|
KH |
50 |
2 |
2 x 10-5 |
0.5 |
(density of water = 103 kg m-3 at 298 K)
This table implies that:
1) solubility of at 308 K is lower than at 298 K
2) The pressure of a 55.5 molal solution of is 250 bar
3) has the highest solubility in water at a given pressure
4) The pressure of a 55.5 molal solution of is 1 bar
p = KHX mol fraction of gas in liquid.
On increasing tamp, ‘K’H increases.
Hence solubility decreases.
Solution: (1)
(1 kJ mol-1 = 83.7 cm-1)
1) 83.7
2) 242.5
3) 145.5
4) 97
[Ti(H2O)6]3+ Ti3+ 3d1 in the octahedral field of ligand.
CFSE = –0.4 Δ0
CFSE = [0.4 * 20300] / 83.7
= 97 kJ mol
Solution: (4)
1) 109
2) 102
3) 119
4) 108
Solution: (1)
Unnilennium 109
Solution: (2)
A student writes general characteristics based on the given mechanism as:
(a) The reaction is favoured by weak nucleophiles.
(b) R𝝝 would be easily formed if the substituents are bulky.
(c) The reaction is accompanied by racemization.
(d) The reaction is favoured by non-polar solvents. Which observations are correct?
1) (a) and (b)
2) (a), (b) and (c)
3) (a) and (c)
4) (b) and (d)
Solution: (2)
Statement (a), (b) & (c) are correct for SN1 reaction mechanism.
1) The diameter of dispersed particles is much smaller than the wavelength of light used.
2) The diameter of dispersed particles is much larger than the wavelength of light used.
3) The refractive index of the dispersed phase is greater than that of the dispersion medium.
4) The diameter of dispersed particles is similar to the wavelength of light used.
Diameter of dispersed particles should not be much smaller than the wavelength of light used.
Refer topic surface chemistry
Solution: (4)
1) CNaCl (T2) > CBaSo4 (T1) for T2 > T1
2) CBaSo4 (T2) > CNaCl (T1) for T2 > T1
3) Ionic mobilities of ions from both salts increase with T.
4) CNaCl >> CBaSo4 at a given T
Ionic
CNaCl >> CBaSo4 at temp ‘T’.
Solution: (4)
1) 3, 3 and 3
2) 4, 2 and 1
3) 2, 4 and 1
4) 4, 2 and 0
P – OH bonds = 4
P = O bonds = 2
P – O – P linkage = 1
4, 2, 1 is the answer.
Solution: (2)
Sol. 47 %
xGlucose = 0.1
mass% of glucose = {[0.1* 180] / [0.1 * 180 + 0.9 * 18]} * 100
= 1800 / [18 + 16.2]
= [1800 / 34.2] %
= 52.63%
= 53%
The mass % of H2O = 47%
Volume strength = [8.9 / 2] * [0.821 * 273] / 1
= 99.73
= 100
Solution: 100
Density = [Z * GMM] / [NA * a3]
2.7 × 103 = [Z * 2.7 * 10-2] / [6.023 * 1023 * (405 * 10-12)]3
Z = 6.023 × 405 × 405 × 405 × 1023–36+3+2
Z = 6.023 × 405 × 405 × 405 × 10–8
Z = 4
FCC
4R = √2 a
R = [405 / 2√2] * 10-12 = 143.21 × 10–12 m
= 143
Solution: 143
Solution: 8
2.303 (RT / F) = 0.06 V; E0AgCl|Ag|Cl- = 0.22 V
Energy of photon = 2.3 – Ecell {for Na}
Energy of photon = 2.25 – Ecell {for K}
Ecell {for ‘Na’} + 0.05 = Ecell {for ‘K’}
0.22 + 0.06 log [H+][Cl–] + 0.05 = 0.22 + 0.06 log [H+] [Cl–]
6 log (10–2) + 5 = 6 log [H+][Cl–]
log (10–12) + log (105) = log {[H+][Cl–]}6
{[H+][Cl–]}6 = 10–7
[H+]12 = 10–7pH = 7 / 12 = 0.58
= 58 × 10–2 = 58
Solution: 58
Video Lessons – September 3 Shift 1 Chemistry
JEE Main 2020 Chemistry Paper With Solutions Shift 1 September 3
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