1 Thermodynamics
The bond energies of C−C, C=C, H−H and C−H linkages are 350, 600, 400 and 410 kJ/mol respectively. The heat of hydrogenation of ethylene is: 1) −170 kJ/mol 2) −400 kJ/mol 3) −260 kJ/mol 4) −460 kJ/mol
Solution
2 General Principles & Process of Isolation of Metals
Leaching of Ag2S is carried out by heating it with a dilute solution of: 1) NaCN only 2) HCl 3) NaOH 4) NaCN in presence of O2
SOLUTION
3 Atomic Structure
In the b-particle emission, the daughter nuclide will be an _______ of the parent.
1) Isotone 2) Isotope or isobar 3) Isobar 4) Isotope
4 Titrations
0.3g of an oxalate salts was dissolved in 100mLsolution. The solution required 90mL of N/20KMnO4 for complete oxidation. The %of oxalate ion in salt is::
1) 3.3%
2) 66%
3) 70%
4) 40%
Meq. Of KMnO4=M eq. of C2O4−2 90×1 / 2=100×NC2O4−2
Mmole of oxalate = $$\displaystyle=\frac{{{9}}}{{{2}\times{2}}}=\frac{{{9}}}{{{4}}}$$
Weight of oxalate =9 ×88×10−3/ 4=22×9×10−3
198×10−3
= $$\displaystyle\%{C}_{{{2}}}{{O}_{{{4}}}^{{-{2}}}}=\frac{{{.198}}}{{{.300}}}\times{100}={66}\%$$
5 Chemical Bonding and Molecular Structure
The number of s and p bonds in Borazole are
1) 12s and 3p
2) 6s and 6p
3) 12s and 12p
4) None
6 General Principles & Process of Isolation of Metals
Thermal decomposition method is used to purify
1) Ti
2) Ni
3) Zr
4) Cr
7 Some Basic Principles of Organic Chemistry
Dipole moment is shown by
1) cis-1, 2-dichloroethene 2) trans-1, 2-dichloroethene 3) trans-1 2-dichloro-2 pentene 4) Both (1) and (3)
SOLUTIONS
C is – 1, 2 – dichloro and trans 1, 2- dichloro – 2 – pentene have dipole moment where as in trans – 1, 2 – dichloro ethene bond moments cancel out hence it has μ = OD. The dipole moment is the vector sum of all the bond.
8 Surface Chemistry
Calculate the surface area in (sq m) of a catalyst that adsorbs 103cm3 of nitrogen reduced to STP per gram in order to form the monolayer. The effective area (in m2) occupied by N2? molecule on the surface is: ( surface area of N2?=1.62×10−15cm2)
1) 4115 m2 2) 2355 m2 3) 4355 m2 4) 435.5 m2
According to the given data,
Number of N2? molecules = Volume(ml) × NA / Volume at STP(in ml)
= (103×6.023×1023) / 22400 = 2.69×1022
Total area covered by N2? = 2.69×1022 × 1.62×10−15 = 435×105 cm2 = 4355 m2.
9 Chemical Thermodynamics
If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the ratio of 1:1:0.5 and ?Hf for the formation of XY is -200 kJ mol-1 . The bond dissociation energy of X2 will be
1) 400 kJ mol-1 2) 300 kJ mol-1 3) 20 kJ mol-1 4) 800 KJmol-1
10 Alcohols Phenols and Ethers
The ionization constant of phenol is more than that of ethanol because:
1) phenoxide ion is a stronger base than ethoxide ion 2) phenoxide ion is stabilized by resonance 3) ethoxide ion is stabilized by resonance 4) phenoxide ion is aromatic while ethoxide ion is aliphatic
The ionization constant of phenol is more than that of ethanol because phenoxide ion is stabilized by resonance. Note: Ionization constant(Ka) of phenol is more than that of ethanol indicates phenol is move acidic than alcohol. Phenol on ionization gives phenoxide ion it is stabilized by resonance
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