KCET 2021 Chemistry Questions with Answers Key Solutions

Solution:

Solution:

Let bond enthalpy of $A _{2}$ be $x , B _{2}$ be $\frac{ x }{2},AB$ be $x$
$A _{2}+ B _{2} \rightarrow 2 AB$
$x +\frac{ x }{2}\,\,\,2x$
$\Delta H _{\text {form }}=\frac{1}{2} A _{2}+\frac{1}{2} B _{2} \rightarrow AB$
$-100=\left(\frac{ x }{2}+\frac{ x }{4}\right)- x$
$x =400$
$B _{2}=\frac{ x }{2}=200 \,kJ$

Solution:

Lesser the steric hindrance, easier and higher the rate of reaction towards $SN ^{2}$

Solution:

$CH _{2}= CH - CH _{2}OH \xrightarrow[\text{Excess} ]{HBr} CH_3 - \underset{\overset{|}{B}r}{CH} - CH_2Br$

Solution:

Solution:

Equilibrium constant is directly proportional to temperature

Solution:

Solution:

Solution:

Solution:

$K _{ a }( HCN )=4 \times 10^{-10} $
$HCN \rightleftharpoons H ^{+}+ CN ^{-}$
Weaker the acid, stronger the conjugate base.

Solution:

Solution:

$\underset{\text{Ethanoic acid}}{CH _{3} COOH} \xrightarrow{\text { LiAlH }_{4} / \text { ether }} \underset{\text{Ethanol}}{CH _{3} CH _{2} OH}$

Solution:

Solution:

Oxidation is addition of oxygen or electronegative element;
Oxidation is removal of hydrogen or electropositive element.

Solution:

Solution:

Solution:

Gabriel phthalimide synthesis not preferred for preparing aniline because amine formation involves nucleophilic substitution of alkyl halides by the anion formed by phthalimide. But aryl halides do not undergo nucleophilic substitution.

Solution:

Clark’s method is for the removal of temporary hardness.

Solution:

Solution:

Deoxyribose and ribose sugars are D-chiral sugars in DNA and RNA.

Solution:

The solubility of alkaline earth metal hydroxides increases with increase in atomic number.

Solution:

Solution:

$MCl + H _{2} SO _{4} \longrightarrow MHSO _{4}+ HCl _{( g )}$

Solution:

$12 : 5$ Membered ring
$20 : 6 $ Membered ring

Solution:

$CCP = N =4$
Octahedral void $= N =1 \times 4=4 \times \frac{1}{4}=1: Be$
Tetrahedral void $=2 N =2 \times 4=2 \times \frac{1}{4}=2: A l$

Solution:

Solution:

$d =\frac{ ZM }{ a ^{3} N _{ A }}$
$6.15=\frac{2 \times M }{\left(3 \times 10^{-8}\right)^{3} \times 6.02 \times 10^{23}}$
$M =\frac{6.15 \times 27 \times 10^{-24} \times 6.02 \times 10^{23}}{2}$
$M =50\, g / mol$

Solution:

Total pressure is $5 atm$. The mole fraction of nitrogen is $0.8$.
Hence, the partial pressure of nitrogen $= P _{ T } \times X =0.8 \times 5=4 atm$
According to Henry's law, $P = KX$
$x$ is the mole fraction
K is the henry's law constant
$P$ is the pressure in atm.
Substitute values in the above expression.
$
X =\frac{4 atm }{1 \times 10^{5} atm }=4 \times 10^{-5}=\frac{ n }{10}
$
$
n =4.0 \times 10^{-4} mol
$

Solution:

Number of moles of $C =\frac{2.4}{12}=0.2$
Number of moles of $H =\frac{1.2 \times 10}{6.0 \times 10}=0.2$
Number of moles of $O =0.2 \,mol$
Empirical formula is $CHO$.

Solution:

option 3

Solution:

$K _{ H }$ is inversely proportional to solubility of gas in liquid.

Solution:

$p _{\text {tot }}= p _{ A }^{0} x _{ A }+ p _{ B }^{0} x _{ B }$
$600=450 x _{ A }+700\left(1- x _{ A }\right)$
$600=450 x _{ A }+700-700 x _{ A }$
$-100=-250 x _{ A }$
$x _{ A }=\frac{100}{250}=\frac{2}{5}$
$=0.4$
$x _{ B }=1- x _{ A }=1-0.4$
$=0.6$

Solution:

Electrode potential is directly proportional to concentration of $Zn ^{2}$ ions.

Solution:

Number of angular nodes $=l$
Number of radial nodes $= n -l-1$
$\therefore $ For $3p$ orbital, angular nodes $=1$
For $3 p$ orbital, radial nodes $=3-1-1=1$

Solution:

$\kappa=\frac{1}{ R }\left(\frac{l}{ a }\right)$
$\therefore \frac{l}{ a }=\kappa \cdot R$
$=0.146 \times 10^{-3} \times 1500$
$=219 \times 10^{3}=0.219\, cm ^{-1}$

Solution:

$E _{\text {cell }}^{0}=\frac{0.059}{ n } \log K _{ C }$
$0.22=\frac{0.059}{2} \log K _{ c }$
$\log K _{ c }=\frac{0.22 \times 2}{0.059}=7.457$
$K _{ C }=2.8 \times 10^{7}$

Solution:

For first-order reaction, half-life $t _{1 / 2}=\frac{0.693}{ k }$
Since half-life does not depend on the concentration of reactants.
According to the given data, the order of reaction with respect to $B$ is one, hence, does not change half-life.

Similarly, an order of reaction with respect to $A$ is also one since the rate increase twice on doubling the concentration.
Hence, total order $=1+1=2$
Second order reaction and its unit is $Lmol ^{-1} s ^{-1}$.

Solution:

Alkaline earth metals have $ns ^{2}$ configuration.
After the removal of $2^{\text {nd }}$ electron, they get stable noble gas configuration.

Solution:

$t =\frac{2.303}{ k } \log \frac{100}{1}$
$=\frac{2.303}{ k } \times 2$
$t =\frac{4.606}{ k }$

Solution:

$r = K [ A ][ B ]^{2}$
When volume is reduced to half, pressure increases by two times.
$r = k [2 A ]^{1}[2 B ]^{2}=2 \times 4=8$ times

Solution:

Based on lowest sum of locants,

4-Ethyl-1-fluoro-2-nitrobenzene

Solution:

Collisions between more than three reactant molecules is rare.

Solution:

Higher the percentage s-character, higher the acidic character.

Solution:

Hardy Schulze Law: The effective ions of the electrolytes in bringing about coagulation are those which carry a charge opposite to that of colloidal particles. Greater the valency of coagulating or fluctuating ion, greater is its power to bring about coagulation.

When colloidal solution is subjected to electrical field, coagulating power should be in the order of $AlCl _{3}> BaCl _{2}> NaCl$ as thier valency is increasing.

Solution:

$\pi$-electrons are referred to as mobile electrons (with respect to graphite).

Solution:

The potential difference between fixed charged layer and the diffused layer having opposite charges is known as zeta potential

Solution:

$NH _{4} NO _{3} \xrightarrow{\Delta}N _{2} O +2 H _{2} O$

Solution:

Thermal stability and bond angle decreases down the group due to increase in bond length.

Solution:

Boiling point is directly proportional to molecular mass and strength of hydrogen bond

Solution:

$XeF _{6}+ H _{2} O \longrightarrow XeOF _{4}+2 HF$

Solution:

$\underset{ P }{ NO } \xrightarrow{[ O ]} \underset{ Q }{ NO _{2}} \xrightarrow{\text { cooling }} \underset{ R }{ N _{2} O _{4}} \xrightarrow{[ H ]} \underset{S}{ N _{2} O _{3}}$

Solution:

$\underset{24}{ Cr }[ Ar ] 3 d ^{5} 4 s ^{1} \,\,\, Cr ^{2+} 3 d ^{4} 4 s ^{0}=4$ unpaired electrons
$Mn [ Ar ] 3 d ^{5} 4 s ^{2} Mn ^{2+} 3 d ^{5} 4 s ^{0}=5$ unpairedelectrons
$Fe [ Ar ] 3 d ^{6} 4 s ^{2} \,\,\,Fe ^{2+} 3 d ^{6} 4 s ^{0}=4$ unpaired electrons
Since both $Cr ^{2+}$ and $Fe ^{2+}$ contain $4$ unpaired electrons, the correct option is $C$.
Option D is also correct because density order given is not correct.

Solution:

Paramagnetism is due to the presence of one or more unpaired electrons

Solution:

From the combined gas equation,
$\frac{p_{1} V_{1}}{T_{1}}=\frac{p_{2} V_{2}}{T_{2}} $
$\frac{p_{1} V_{1}}{T_{1}}=\frac{\frac{1}{2} p_{1} V_{2}}{2 T_{1}}$
$V_{1}=\frac{1}{4} V_{2}$
$V_{2}=4 V_{1}$

Solution:

$Sc =21, Sc ^{3+}=[ Ar ] 3 d ^{0} 4 s ^{0}=$ no unpaired electrons
$Ti =22, Ti ^{3+}=[ Ar ] 3 d ^{1} 4 s ^{0}=1$ unpaired electron
$Ti ^{4+}=[ Ar ] 3 d ^{\circ} 4 s ^{\circ}$ no unpaired electrons
$Ni =28, Ni ^{2+}=[ Ar ] 3 d ^{8} 4 s ^{2}=2$ unpaired electrons
$Cu =29, Cu ^{+}=[ Ar ] 3 d ^{10} 4 s ^{0}=$ no unpaired electrons
$Mn =25, Mn ^{2+}=[ Ar ] 3 d ^{5} 4 s ^{2}=5$ unpaired electrons

Solution:

Solution:

Due to free radical mechanism (secondary free radical)

Solution:

Pentaamminecarbonatocobalt(III) chloride

Solution:

Complex compounds having only one type of ligands are homoleptic complexes

Solution:

Wavelength of light absorbed is inversely proportional to strength of the ligand.