I. dilute hydrochloric acid and
II. aqueous sodium hydroxide.
The ratio of the volumes of H2 evolved in these two reactions is:
Zn + 2NaOH Na2ZnO2 + H2
Zn + 2HCl ZnCl2 + H2
So, the ratio of volume of H2 released in both the cases is 1:1.
Cr(OH)3 (s) Cr3+(aq) + 3OH–(aq)
1-S S 3S
Ksp = 27S4
6×10-31 = 27S4
S = [(6/27)×10-31]1/4[OH–] = 3S = 3×[(6/27)×10-31]1/4
= (18×10-31)1/4 M
a) Lithium has the highest hydration enthalpy among the alkali metals.
b) Lithium chloride is insoluble in pyridine.
c) Lithium cannot form ethynide upon its reaction with ethyne.
d) Both lithium and magnesium react slowly with H2O.
a) (a), (b) and (d) only
b) (b) and (c) only
c) (a), (c) and (d) only
d) (a) and (d) only
Only LiCl amongst the first group chlorides dissolve in pyridine because the solvation energy of lithium is higher than the other salts of the same group. Lithium does not react with ethyne to form ethynilide due to its small size and high polarizability. Lithium and Magnesium both have very small sizes and very high ionization potentials so, they react slowly with water. Amongst all the alkali metals, Li has the smallest size hence, the hydration energy for Li is maximum.
a) 1 and 2
b) 1 and 0.5
c) 1 and 1
d) 2 and 0.5
The given data for ionization energies clearly shows that IE2 ≫ IE1. So, the element belongs to the first group. Therefore, we can say that this element will be monovalent and hence forms a monoacidic base of the type MOH.
MOH + HCl MCl + H2O
2MOH + H2SO4 M2SO4 + 2H2O
So, from the above equation we can say that, 1 mole of metal hydroxide requires 1 mole of HCl and 0.5 mole of H2SO4.
Let us assume the equation to be A ⇌ B,
Number of particles of A = 6
Number of particles of B = 11
K= 11/6 ≈ 2
III. Na3[Fe(C2O4)3] (0 > P)
Complex (I) has the central metal ion as Fe2+ with strong field ligands.
Configuration of Fe2+ = [Ar] 3d6
Strong field ligands will pair up all the electrons and hence the magnetic moment will be zero.
Complex (II) has the central metal ion as Cr2+ with weak field ligands.
Configuration of Cr2+ = [Ar] 3d4
As weak field ligands are present, pairing does not take place. There will be 4 unpaired electrons and hence the magnetic moment = √24 B.M.
Complex (III) has the central metal ion as Co2+ with weak field ligands.
Configuration of Co2+ = [Ar] 3d7
As weak field ligands are present no pairing can occur. There will be 3 unpaired electrons and hence the magnetic moment = √15 B.M.
Complex (IV) has the central metal ion as Fe3+ with strong field ligands.
Configuration of Fe3+ =[Ar] 3d5
Strong field ligands will pair up the electrons but as we have a [Ar] 3d5 configuration, one electron will remain unpaired and hence the magnetic moment will be √3 B.M.
a) S is a function of temperature but S is not a function of temperature.
b) Both S and S are functions of temperature.
c) Both S and S are not functions of temperature.
d) S is not a function of temperature but S is a function of temperature.
Entropy is a function of temperature, at any temperature, the entropy can be given as:
Change in entropy is also a function of temperature, at any temperature, the entropy change can be given as:
a) Boron nitride, MeBr
b) Diborane, MeMgBr
c) Borazine, MeBr
d) Borazine, MeMgBr
H3N3B3Cl3 + LiBH4 B3N3H6 + LiCl + BCl3
B3N3H3Cl3 + 3CH3MgBr B3N3H3(CH3)3 + 3MgBrCl
So, we can say that,
B is B3N3H6 and C is CH3MgBr
As H2,O2 and CO gets adsorbed on the surface of charcoal, the pressure decreases. So, option (a) and (d) can be eliminated. After some time, as almost all the surface sites are occupied, the pressure becomes constant.
a) cis only
b) trans only
c) meridional and trans
d) cis and trans
In cis-isomer, similar ligands are at an angle of 900.
a) distilled water
b) sea water
c) saline water used for intra venous injection
d) water from a well
In distilled water there are no ions present except H+ and OH– ions, both of which are immensely minute in concentration, that renders their collective conductivity negligible.
Benzene (C6H6) has 6 sp2 hybridized carbons. Each carbon has 3 -bonds and 1 -bond. 3 -bonds means that there are 3 sp2 hybrid orbitals for each carbon. Hence, the total number of sp2 hybrid orbitals is 18.
a) Cl- CH=CH2
There is extended conjugation present in option (d), which will reduce the length of C-Cl bond to the greatest extent which can be represented as follows:
a) for the photochemical breakdown of waste present in 1m3 volume of a water body.
b) by anaerobic bacteria to break-down inorganic waste present in a water body.
c) by bacteria to break-down organic waste in a certain volume of water sample.
d) for sustaining life in a water body.
Biochemical oxygen demand (BOD) is the amount of dissolved oxygen used by microorganisms in the biological process of metabolizing organic matter in water.
c) Nylon 6,6
A, B and C are respectively
a) A = Lactose B = Glucose C = Albumin
b) A = Lactose B = Glucose C = Alanine
c) A = Lactose B = Fructose C = Alanine
d) A = Glucose B = Sucrose C = Albumin
Lactose, glucose and fructose gives positive Molisch’s test.
Glucose gives positive Barfoed’s test whereas sucrose gives a negative for Barfoed’s test.
Albumin gives positive for Biuret test whereas alanine gives a negative Biuret test.
a) I > II > III > IV
b) IV > III > I > II
c) II > I > III > IV
d) IV > I > II > III
The basicity of the compound depends on the availability of the lone pairs.
In compound IV, Nitrogen is sp3 hybridized.
In compound III, Nitrogen is sp2 hybridized and the lone pairs are not involved in resonance.
In compound I, Nitrogen is sp2 hybridized and the lone pairs are involved in resonance.
In compound II, Nitrogen is sp2 hybridized and the lone pairs are involved in resonance such that, they are contributing to the aromaticity of the ring. From the above points we can conclude that the basicity order should be IV > III > I > II.
The generation time can be utilized to get an indication of the rate ratio. Let the amount generated be (x).
Rate = Amount generated/Time taken
Rate300 K = (x)/60
Rate400 K =(x)/40
Rate 300K/Rate 400K = 40/60
For the same concentration (which is applicable here), the rate ratio can also be equal to the ratio of rate constants.
Ea = 0.4 × 8.3 × 1200 = 3984 J/mol = 3.98 kJ/mol
Ppm = (WSolute/W Solution)×100
Using the density of the solution and its volume (1L = 1000 mL = 1000 cm3), the weight of the solution can be calculated.
Wsolution = 1.03 × 1000 = 1030 g
Thus, ppm = ((10.3 x 10 -3g)/1030 g) × 100
Kf = 2
Molality, ‘m’ = 0.5
Tf = Kf × m
= (0.5 × 2) = 1
So, the initial temperature now becomes 272 K. Further using the given value of moles and initial volume of the gas and the calculated initial temperature value, we can find out the initial pressure of the ideal gas contained inside the piston.
Pgas = nRT/V1
= (0.1)(0.08)(272) = 2.176 atm
Now, on releasing the piston against an external pressure of 1 atm, the gas will expand until the final pressure of the gas, i.e. P2 becomes equal to 1 atm. During this expansion, since no reaction is happening and the temperature of the gas is not changing as well, the boyle’s law relation can be applied.
P1V1 = P2V2
2.176 × 1 = 1 × V2
V2 = 2.716
The mass percentage of carbon in A is:
Compound A is CH3(CO)CH2CH3 (C4H8O)
The percentage of carbon in compound A by weight is WCarbon/WCompound = 12×4/72 = 66.67