Class 12th Chemistry 2017 Set2 Delhi Board Paper Solution
Give reasons:
CH3 NH2 is more basic than C6H5NH2.
Due to resonance the lone pair of electrons on the nitrogen in aniline get delocalised over the benezene ring and thus, is less easily available for protonation. Hence, aniline is weaker base than methyle amine.
Popular Questions of Class 12th chemistry
- Q:-
How the following conversions can be carried out?
(i) Propene to propan-1-ol
(ii) Ethanol to but-1-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4-dimethylhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide
- Q:-
Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid
(iv) Benzoic acid and Ethyl benzoate
(v) Pentan-2-one and Pentan-3-one
(vi) Benzaldehyde and Acetophenone
(vii) Ethanal and Propanal
- Q:-
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
- Q:-
A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL-1, then what shall be the molarity of the solution?
- Q:-
Henry's law constant for CO2 in water is 1.67 x 108Pa at 298 K. Calculate the quantity of CO2in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.
- Q:-
Calculate the mass of a non-volatile solute (molar mass 40 g mol-1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
- Q:-
The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
- Q:-
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
- Q:-
How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?
- Q:-
If NaCl is doped with 10-3mol % of SrCl2, what is the concentration of cation vacancies?
Recently Viewed Questions of Class 12th chemistry
- Q:-
Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co(H2O)(CN)(en)2]2+
(ii) [CoBr2(en)2]+
(iii) [PtCl4]2-
(iv) K3[Fe(CN)6]
(v) [Cr(NH3)3Cl3]
- Q:-
Q1 : Write down the electronic configuration of:
(i) Cr3+ (iii) Cu+ (v) Co2+ (vii) Mn2+
(ii) Pm3+ (iv) Ce4+ (vi) Lu2+ (viii) Th4+
- Q:-
Give the IUPAC names of the following compounds:
(i) PhCH2CH2COOH (ii) (CH3)2C=CHCOOH
(iii)
(iv) 
- Q:-
Mention the factors that affect the rate of a chemical reaction.
- Q:-
Classify the following solids in different categories based on the nature of intermolecular forces operating in them:
Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide. - Q:-
Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
- Q:-
Explain why Cu+ ion is not stable in aqueous solutions?
- Q:-
Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.
- Q:-
What is shape selective catalysis?
- Q:-
Give reason for the higher boiling point of ethanol in comparison to methoxymethane.