KCET 2022 Chemistry Questions with Answers Key Solutions

Solution:

$t _{99.9 \%}=10 t _{50 \%}$
$=10 \times 45\, min =450\, min =7.5$ hours

Solution:

$2^{\text {nd }}$ order reaction

Solution:

$\lambda_{ m }=\frac{1000 \,k }{ C }=\frac{1000 \times 6.3 \times 10^{-2}}{0.1}$
$=630 \,ohm ^{-1}\, cm ^{2} \,mol ^{-1}$

Solution:

$t _{1 / 2} \propto \frac{1}{ a ^{ n -1}}$

Solution:

$t _{1 / 2} \propto \frac{1}{ a ^{ n -1}} ; n =6$

Solution:

$n =4$
$Fe ^{+2}=3 d ^{6}$

Solution:

$\left( NH _{4}\right)_{2} Cr _{2} O _{7} \xrightarrow{\text { heat }} Cr _{2} O _{3}+4 H _{2} O + N _{2}$
$NH _{4} NO _{2} \xrightarrow{\Delta} 2 H _{2} O + N _{2}$

Solution:

$\left[ Pt \left( NH _{3}\right)_{6}\right] Cl _{4} \longrightarrow\left[ Pt \left( NH _{3}\right)_{6}\right]^{+4}+4 Cl ^{-}$
Five ions are produced

Solution:

Reactivity order of $Sn ^{1}$ reaction is $3^{\circ}>2^{\circ}>1^{\circ}$

Solution:

$\left[Pt\left( NH _{3}\right)_{2} Cl _{2}\right]=$ cis-platin

Solution:

$\Delta_{ t }=\frac{4}{9} \Delta_{0}$
$=\frac{4}{9} \times 18000\, cm ^{-1}$
$=8000\, cm ^{-1}$

Solution:

$CH _{3}- CH _{2}- OH \xrightarrow[443\,K]{Conc. H_2SO_4} CH _{2}= CH _{2}$

Solution:

Solution:

on solvolysis give more stable benzyl carbocation

Solution:

$R - X + AgCN \rightarrow \underset{(A)}{R - NC} + AgX$
$R - X + KCN \rightarrow R \underset{( B )}{- CN} + KX $

Solution:

Phenols gives characteristic colour with $FeCl _{3}$

Meta-derivative of phenol only gives tribromo derivative

Solution:

Solution:

Carbylamine test for primary amines the resulting isocyanide
$CH _{3}- NH _{2} \xrightarrow[NaOH]{CHCl_3} CH _{3} NC$

Solution:

Carboxylic acid with alpha hydrogen undergoes $HVZ$ reaction

Solution:

Solution:

Solution:

Solution:

Solution:

$W =-2.303\, nRT \log V _{2} / V _{1}$

Solution:

$n _{ C _{2} H _{5} OH }=\frac{414}{46}=9$
$n _{ H _{2} O }=\frac{18}{18}=1$
$X _{ H _{2} O }=\frac{1}{10}=0.1$

Solution:

$\lambda=\frac{ h }{ mv }=\frac{ h }{ p }$
$p =\frac{ h }{\lambda}=\frac{6.6 \times 10^{-34}}{2.2 \times 10^{-11}}=3 \times 10^{-23}$

Solution:

Its low solubility in water makes it of biological and geo-chemical importance. It form carbonic acid with water which dissociates to give $HCO _{3}^{-}$ ions. $H _{2} CO _{3} / HCO _{3}^{-}$buffer system helps to maintain $pH$ of blood between $7.26-7.42$

Solution:

$ K _{ C } \frac{\left[ H _{2}\right]\left[ I _{2}\right]}{[ HI ]^{2}} $
$ K _{ C }= \frac{\frac{0.25 \times 0.25}{2}}{\frac{0.5 \times 0.5}{2}}$
$=\frac{1}{4}=0.25 $

Solution:

$\Delta_{ m }=\frac{ k \times 1000}{ M }$
Lower the molarity higher the molar conductivity

Solution:

$\Delta T _{ b }= K _{ b } . m . i $
$\Rightarrow 0.1= K _{ b } \times \frac{1.8}{180} \times \frac{1000}{100} \times 1$
$K _{ b }=1$

Solution:

$\pi= C . R . T =\frac{ w _{2}}{ M _{2}} \frac{1000}{ V ( m \ell)} \times R . T$
$\Rightarrow \frac{3}{180} \times \frac{1000}{60} \times 0.0821 \times 288=6.568 \,atm$