NEET 2022 Chemistry Questions with Answers Key Solutions

Solution:

Weak acid $\left( CH _{3} COOH \right)$ and salt of weak acid-strong base $\left( CH _{3} COONa \right)$ form an acidic buffer.
Sodium acetate $\left( CH _{3} COONa \right)=0.10 \,M$;
Acetic acid $\left( CH _{3} COOH \right)=0.01 \,M$;
$pH$ of acidic buffer solution is given by
$pH = pK _{ a }+\log \frac{[\text { Salt }]}{[\text { Acid }]} $
$=4.57+\log \left(\frac{0.1}{0.01}\right)$
$=5.57$

Solution:

Dalton's law of partial pressure :
Partial pressure of gas $=$ mole fraction of gas in gaseous mixture $\times$ Total pressure of gaseous mixture.
$p _{1} = X _{1} p $
$p _{2} = X _{2} p $
$ p _{3} = X _{3} p$
Total pressure,
$p = p _{1}+ p _{2}+ p _{3}$

Solution:

Which is incorrect statement regarding enzymes
(1) Like chemical catalysts enzymes reduce the activation energy of bio process $\Rightarrow$ This is correct statement.
(2) Enzymes are polysaccharides $\Rightarrow$ This is incorrect statement because enzymes are protein in nature
(3) Enzymes are very specific for a particular reaction and substrate $\Rightarrow$ This is correct statement.
(4) Enzymes are biocatalyst $\Rightarrow$ This is correct statement.

Solution:

$Li$ - Electrochemical cells
$Na$ - Coolant in fast breeder reactors
$KOH$ - absorbent for $CO _{2}$
$Cs$ - Photoelectric cell.

Solution:

Solution:

Solution:

$KO _{2}$
$K ^{+} O _{2}^{-}\left( O _{2}^{-}-\right.$superoxide ion$)$

Solution:

Solution:

SRP : $ E _{ Zn ^{2+} / Zn }^{\circ}< E _{ Fe ^{2+} / Fe }^{\circ}< E _{ Cu ^{2+} / Cu }^{\circ}< E _{ Ag ^{+} / Ag }^{\circ}$
Reactivity order : $Zn > Fe > Cu > Ag$
In case of displacement reaction, more reactive metals (lower SRP) can displace less reactive metals (higher SRP) from their salt solution.
$CuSO _{4(\text { aq. })}+2 Ag _{( s )} \rightarrow Cu _{( s )}+ Ag _{2} SO _{4(\text { aq. })}$
Option (3)
Reaction is not possible
as $Ag$ is less reactive metal compare to $Cu$.

Solution:

Acidic strength of phenolic group increases due to electron withdrawing groups.
Order of acidic strength

Solution:

Solution:

(i) Statement-1 is correct because in point defects of ionic solid electrical neutrality is essential condition (given question is example of metal deficiency defect)
(ii) Statement-2 is correct because In Frenkel defect cation dislocate from lattice site to interstitial position.
(iii) Both statement are correct but statement-2 is not correct explanation of statement-1

Solution:

Boiling point of comparable molecular mass molecules

Solution:

Interhalogen compound group $17^{\text {th }}$
$ICl$ is more reactive due to polar bonds.
From NCERT - X-X' bond is weaker than $X - X$ bond except $F _{2}$

Solution:

Solution:

Electron deficient hydride $\rightarrow$ Less than $8 e^{-}\left( B _{2} H _{6}\right)$
Electron precise hydride $\rightarrow$ having $8 e^{-}$without l.p. $\left( GeH _{4}\right)$
Electron rich hydride $\rightarrow$ having $8 e^{-}$with l.p. $(HF)$

Solution:

${ }_{64} Gd =[ Xe ] 6 s ^{2} 4 f ^{7} 5 d ^{1} $
$Gd ^{+2} =[ Xe ] 4 f ^{7} 5 d ^{1}$
After losing $5 d$ electron $4 f$ has maximum exchange energy so Gd has low value of Third Ionisation energy

Solution:

According to Hardy Schulze Rule statement 1 is correct. (Generally, the greater the valence of the flocculating ion added, the greater is its power to cause precipitation)
According to Hardy Schulze Rule statement 2 is incorrect

Solution:

Antacid - Cimetidine
Antihistamine - Seldane
Analgesic - Morphine
Antimicrobials - Salvarsan

Solution:

Kjeldahl's method is not applicable to the compounds containing nitrogen having nitro and azo group and nitrogen present in the ring (pyridine), as nitrogen of these compounds does not change to ammonium sulphate under these conditions.

Solution:

In diamond each carbon is bonded with four other carbon atoms. So hybridisation of carbon atom is $sp ^{3}$.
In graphite each carbon is bonded with three other carbon atoms. So hybridisation of carbon atom is $sp ^{2}$.

Solution:

IUPAC
$\left[ Ag \left( H _{2} O \right)_{2}\right]\left[ Ag ( CN )_{2}\right]$
Coordination number $=2$,
Oxidation state $= Ag ^{+1}$
Diaquasilver(I) dicyanidoargentate(I)

Solution:

In $P - V$ graph area under the curve represent magnitude of work.
As it is maximum in graph- $1$

Solution:

$m =\frac{\text { Moles of solute }}{\text { Weight of solvent }( g )} \times 1000$
$1=\frac{0.5}{\text { Weight of solvent }( g )} \times 1000$
Weight of solvent $( g )=500\, g$

Solution:

Solution:

$CaCO _{3( s )}+2 HCl _{\text {(aq.) }} \rightarrow CaCl _{2 \text { (aq.) }}+ CO _{2( a )}+ H _{2} O _{(0)}$
no. of moles of $CaCO _{3}$ (pure) $=\frac{1}{2} \times$ mole of $HCl$
$ \text { [Mole }=\text { molarity } \times \text { volume(in ltr. })]$
$=\frac{1}{2} \times 0.5 \times \frac{50}{1000}=0.0125$
weight of $CaCO _{3}$ (pure) $=$ mole $\times$ mol. wt
$=0.0125 \times 100=1.25\,g$
$\%$ purity $=\frac{\text { wt. of pure substance }}{\text { wt. of impure sample }} \times 100$
$95=\frac{1.25}{\text { wt. of impure sample }} \times 100$
wt. of impure sample $=\frac{1.25 \times 100}{95}=1.32\, g$

Solution:

$O _{2}^{+}$ion is having $15 $ electrons, so it contain one unpaired electron. Hence it is paramagnetic in nature.

Solution:

Solution:

Solution:

(I) curve is suitable for zero order if $y=$ rate and $x =$ concentration because in case of zero order reaction rate is constant and does not depend on conc $^{ n }$.
(II) curve is suitable for first order if $y=t_{1 / 2}$ and $x=$ conc $^{n}$ because in case of first order $t_{1 / 2}$ does not depend on conc $^{ n }$.

Solution:

Enantiomers are non-superimposable mirror images of each other.

Solution:

IUPAC nomenclature
$119 \rightarrow \text { Ununennium } \rightarrow \text { Uue }$

Solution:

Thermosetting polymers are NOT reusable.

Solution:

Hydrides of group $16^{\text {th }}$
$\underset{\text{H-bond}}{H _{2} O} \,\,\, \underbrace{ H _{2} S \quad H _{2} Se \quad H _{2} Te }_{ H \text {-bond }}$
B.P. $\rightarrow H _{2} S < H _{2} Se < H _{2} Te < H _{2} O$

Solution:

$XeF _{2}$ has the maximum number of lone pairs of electrons.
Number of lone pair of electrons for $XeF _{2}$ is 3 .

Solution:

A $\rightarrow$ Products
Initial conc. $A _{ o }=0.1 \,M$
Conc. After $5 \min A_{t}=0.001\, M$
$t =5 \min$.
For first order reaction
$K =\frac{2.303}{ t } \log \left(\frac{ A _{ o }}{ A _{ t }}\right)$
$=\frac{2.303}{5} \log \left(\frac{0.1}{0.001}\right) $
$K =0.9212 \min ^{-1}$

Solution:

Solution:

Solution:

Solution:

$3 O _{2}( g ) \rightleftharpoons 2 O _{3}( g )$
$K _{ c }=\frac{\left[ O _{3}\right]^{2}}{\left[ O _{2}\right]^{3}}$
$3 \times 10^{-59}=\frac{\left[ O _{3}\right]^{2}}{\left(4 \times 10^{-2}\right)^{3}}$
${\left[ O _{3}\right]^{2}=3 \times 10^{-59} \times 64 \times 10^{-6} }$
$=19.2 \times 10^{-64}$
$=4.38 \times 10^{-32} M$

Solution:

$\frac{\left( r _{3}\right)_{ L ^{+2}}}{\left( r _{2}\right)_{ He ^{+}}}=\frac{ n _{1}^{2}}{ n _{2}^{2}} \times \frac{ Z _{2}}{ Z _{1}}$
$\frac{\left( r _{3}\right)_{ L ^{+2}}}{105.8\, pm }=\frac{3 \times 3}{2 \times 2} \times \frac{2}{3}$
$\left( r _{3}\right)_{ Li ^{+2}}=158.7\, pm$

Solution:

The presence of particulate matter in polluted air catalyses the oxidation of sulphurdioxide to sulphur trioxide.
$2 SO _{2}( g )+ O _{2}( g ) \rightarrow 2 SO _{3}( g )$
The reaction can also be promoted by ozone and hydrogen peroxide.
$SO _{2}( g )+ O _{3}( g ) \rightarrow SO _{3}( g )+ O _{2}( g ) $
$ SO _{2}( g )+ H _{2} O _{2}( l ) \rightarrow H _{2} SO _{4}( aq )$

Solution:

$V =10 L $
$ W _{ O _{2}}=64 \,g$
$T =2{ }^{\circ} C$
$ n _{ O _{2}}=2$
$R =0.083 . L$ bar $K ^{-1} \,mol ^{-1}$
Ideal gas equation $PV = nRT$
$P =\frac{2 \times 0.0831 \times 300}{10}$
$P =4.9$ bar

Solution:

Solution:

$1^{\circ}, 2^{\circ}, 3^{\circ}$ Alcohol are distinguished by Lucas test on the basis of the time taken for turbidity to appear

Reactivity of alcohol towards Lucas reagent $\Rightarrow 3^{\circ}>2^{\circ}>1^{\circ}$ Alcohol

Solution:

(A) $\left[ Ni \left( H _{2} O \right)_{2}( en )_{2}\right]^{2+}$
(B) $\left[ Ni \left( H _{2} O \right)_{4}( en )\right]^{2+}$
(C) $\left[ Ni ( en )_{3}\right]^{2+}$
en is SFL (strong field ligand)
As the number of en (strong ligand) increase splitting also increases.
So, $\Delta_{0}$ increases.
i.e. maximum energy will be absorbed in case of option $C$.
So the order is $C >A >B$

Solution:

$d =\frac{ Z \times M }{ N _{ A } \times a ^{3}}$
$8.92=\frac{4 \times M }{6.022 \times 10^{23} \times\left(3.608 \times 10^{-8}\right)^{3}}$
$M =\frac{8.92 \times 6.022 \times 10^{23}}{4} \times 46.96 \times 10^{-24}$
$M =63.1\, g / mol$ (Molar Atomic Mass)
$M =63.1 \,u$ (Atomic Mass)

Solution:

$Ni ( s )+2 Ag ^{+}(0.001 M ) \rightarrow Ni ^{+2}(0.001 M )+2 Ag ( s )$
$E _{\text {cell }}= E _{\text {cell }}^{\circ}-\frac{0.059}{ n } \log \frac{\left[ Ni ^{+2}\right]^{1}}{\left[ Ag ^{+}\right]^{2}}$
$E _{\text {cell }}=10.5-\frac{0.059}{2} \log \frac{10^{-3}}{\left(10^{-3}\right)^{2}}$
$=10.5-\frac{0.059}{2} \log 10^{+3}$
$=10.5-\frac{0.059}{2} \times 3$
$=10.4115 \,V$

Solution:

Haematite $Fe _{2} O _{3}$
Magnetite $Fe _{3} O _{4}$
Calamine $ZnCO _{3}$
Kaolinite $\left[ Al _{2}( OH )_{4} Si _{2} O _{5}\right]$

Solution: