Given: Na2 SO4. ⎯⎯→ Na+ = 92 gm
To find: Molality of Na+ per kg of water
Formula: Molality per kg = (Wt of solute) / (molecular wt of solute wt of solvent(kg)) = 92/23 = 4
a) (x/m) ∝ (P)1/2
b) (x/m) ∝ (P)
c) (x/m) ∝ (P)2
d) (x/m) ∝ (P)0
According to Freundlich isotherm
(x/m) ∝ (P)1/n
Where, 1/n implies 0 to 1
(x/m) = k(P) 1/n
Slope= 1/n = (From graph Slope= tan 2/4)
2/4 n = 1/ n
(Given that P Kb of NH4OH is 4.7)
H2SO4+2 NH4OH → (NH4)2SO4+ H2O
20 × 0.1 ; 30 × 0.2
It is a complete Neutralisation ,r x n So Buffer formula is applied
POH = Pkb + log10(Salt/Base)
= 4.7 + log10(4/2)
= 4.7 + log10(2)
P OH = 5
PH = 14 -5 = 9
PH = 9
a) On increasing temperature value of KH increases
b) Value of KH increases solubility of gas increases
c) Value of KH for two different gases at same temperature is not same
d) None of these
According to Henry’s law
P = KH x Solubility
P = Partial pressure of gas
KH = Henry’s constant
Solubility ∝ 1/KH (KH is different for different gas)
According to this expression, if the solubility of gas increases the value of KH decreases.
Rate (M min-1 )
6.93 x 10-3
6.93 x 10-3
1.386 x 10-2
Time when concentration of A becomes half
Rate = k[A] x [B] y
6.93 x 10-3 = k(0.1)x (0.20)y… (1)
6.93 x 10-3 = k(0.1)x (0.25)y… (2)
1.386 x 10-2 = k(0.2)x (0.30)y… (3)
Divide equation (1) by (2)
1= (0.20/0.25)y y = 0
Divide equation (1) by (3)
(1/2) = (1/2)x x 1
From equation (i) and (iii) we get x =1, so it is first order with respect to A.
6.93 x 10-3 = k(0.1 )
k = 6.93 x 10-2 min-1
t 1/2 = 0.693/k
t 1/2 = 0.693/(6.93 x 10-2) =10
a) 30.3 g
b) 15.5 g
c) 7.6 g
d) 60.6 g
Charge = 0.05 F
Amount of PbSO4 precipitated = W
Molar mass of PbSO4 = 303 g/mol
According to Faraday’s 1st law of electrolysis
1 st method
W = E x Q
E = M/2; Q= 0.05
W = (303 x 0.05)/2 = 7.6 g
2 nd method:
Pb(s) + SO4 2- ⎯⎯→ PbSO4 + 2e–
For 2f current passed, PbSO4 deposited = 303 g/mol
For 0.05 F current passed, PbSO4 deposited = W
W = (303x 0.05)/2= 7.6 g
a) Slope = −RH
b) Slope = RH
c) Intercept = RH
d) Graph is non-linear
Transition state = n = 8 to n = nf
Graph (ῡ vs (1/nf2 ))
ῡ = RH z2[(1/n12) – (1/n22)] (Z = 1 for hydrogen emission spectrum)
ῡ = RH 12[(1/nf2) – (1/82)]
Comparing it with y = mc2
Slope = RH
Given: Mix – Gas A + Gas B
Gas A = 0.5 mole and Gas B = x mole
Total pressure = 200 pa
T = 1000 k V = 10 m3
x = ?
Pv = nRT
200 x 10 = (0.5+ x )R x 1000
2 = (0.5 + x) R
(2/R) -0.5 = x
(2/R) – (1/2) = x (4-R/2R) = x
a) Deutrium, Protium
b) Deutrium, Tritium
c) Deutrium, Tritium, Protium
Protium, Deutrium and Tritium
a) P > Q > R
b) Q > P > R
c) R > P > Q
d) R > Q > P
This nitrogen contains 3σ bond and its hybridisation is SP2 . Since the lone pair is delocalised inside the ring it doesn’t take part in hybridisation and make this compound to be aromatic in nature
It involves Electrophilic addition of alkenes, followed by Nucleophilic substitution mechanism.
a) P > Q > R > S
b) R > S > P > Q
c) R > P > S > Q
d) R > S > Q > P
Based on the order of – I effect.
More the -I effect greater its acidic strength.
-NO2 >CN> F >Cl
Hence the correct order is R > S > P > Q
a) Fe = 0.2 ppm
b) Cu = 2 ppm
c) Mn = 0.5 ppm
d) Zn = 0.05 ppm
Fact from NCERT
Presence of Mn ≥ 0.05 ppm concentration makes water unsuitable for drinking
d) CH (CN)3
Order lies with –I and –M effect. -I and –M effect in CN is more when compared to Cl, I and Br. After losing H+ the negative ion on the C will be delocalized to CN making it more stable and acidic.
Ans: CH (CN)3
•It depends on the polarization power of cation.
• The atom which is having larger size will have lesser polarization power and does not have water of crystallization.
• Ba 2+ ion is larger in size in comparison to Ca 2+ Mg 2+ and Sr 2+ ion.
• Ba(NO3)2 is the correct option.
c) Copper pyrite
d) None of these
Malachite – Copper ore CuCO3 Cu(OH)2 .
Azurite – Copper ore 2CuCO3 Cu(OH)2
Copper pyrite – Cu+ Fe CuFes2 – Copper + Iron ore
a) Li2+ stable Li2−unstable
b) Li2+ unstable Li2−unstable
c) Li2+ unstable Li2− unstable
d) Li2+ stable Li2– stable
Li2+ → 5e –
Bond order = (Nb –NA)/2 = (3-2)/2 =1/2
Li2– → 7e –
Bond order = (Nb –NA)/2 = (4-3)/2 =1/2
Both Li2+, + Li2− have same bond order. Unlikely 0.5 bond order does not exist so both Li2+ and Li2− are unstable.
If option were given as
1) Li2+ > Li2–
2) Li2+ < Li2–
3) Li2+ = Li2–
4) None of these
We can consider if same B.O is present, the species which is having lesser number of electrons present in antibonding orbital will be more stable so Li2+ > Li2− .
a) electronegativity and atomic radius
b) electronegativity and electro gain enthalpy
c) atomic radius and electronegativity
d) electro gain enthalpy and electronegativity
On moving down, the group electro negativity decreases. On moving down, the group atomic shell increases there by atomic radius increases.
w = – nRT ln (vf/vi) → constant is given same in all case.
Take magnitude forW.
T2 > T1
Lines cannot intersect and Intercept will be negative.
A) Thermally resistant and have low dielectric constant
B) Resistant towards oxidation and used in grease
D) Hydrophobic in nature
a) A & B
b) A, B & C
c) B, C, & D
d) A, B, C & D
All, A, B, C and D are true for a silicone polymer
Piezoelectric → these are the material that produce electrical potential when pressure is applied on parallel and perpendicular phases.
EX – Quarts.
a) Inert pair effect
b) Lanthanoid contraction
c) Diagonal relationship
d) None of these
Due to inert pair effect thallium exist both 1+ and +3 oxidation state. But Thallium is stable in +1 oxidation state.
a) 5.92 BM
b) 6.92 BM
c) 4.89 BM
d) 3.87 BM
In Transition metal complex maximum number of unpaired electron possible is 5 and it will be present in d sub shell
Formula → =
n no of impaired electron
A) Chloroxylenol P) Carbylamines
B) Penicillin Q) Baeyer’s reagent
C) Sulpha Pyridine R) FeCl3 test
D) Norethindrone S) Sodium hydrogen sulphate
a) A → R, B → P, C → S, D → Q
b) A → S, B → R, C → P, D → Q
c) A → R, B → S, C → P, D → Q
d) A → Q, B → R, C → P, D → S
Penicillin contains COOH group – respond to sodium hydrogen sulphate test.
Chloroxylenol contains OH group – respond to neutral FeCl3 test
Sulpha pyridine NH2 group – respond to carbylamines test
Norethindrone C ≡ CH – respond to Bayers test
a) Ph − CH − Cl Ph− CH2− NC
b) PhCH2OH Ph− CH2 − CN
c) PhCH2 Cl Ph− CH2− CN
d) Ph − OH Ph− CH2− CN
c) R COOH
Classical reduction ⎯⎯→ Aldehyde is formed.
Lysine, Aspartic acid, Arginine, Glycine.
a) lys > Arg > Gly > Asp
b) Arg > Lys > Asp > Gly
c) Gly > Asp > Arg > Lys
d) Arg > Lys > Gly > Asp
It involves nucleophilic substitution reaction (SN2 ) followed by oxidation with oxidising agent and removal of water molecule.
a) (Δ0)A< (Δ0)B
b) The crystal field splitting parameter can be measured by wavelengths of complementary colors for (A) and (B) respectively.
c) Both are paramagnetic with three unpaired electrons each.
d) The crystal field splitting parameter can be measured by wavelength of yellow and violet colors for (A) and (B) respectively.
The crystal field splitting parameter can’t be measured by wavelength of yellow and violet colours for (A) & (B) respectively.