iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 28th July Evening Shift
A sample of $0.125 \mathrm{~g}$ of an organic compound when analyzed by Duma's method yields $22.78 \mathrm{~mL}$ of nitrogen gas collected over $\mathrm{KOH}$ solution at $280 \mathrm{~K}$ and $759 \mathrm{~mm}\, \mathrm{Hg}$. The percentage of nitrogen in the given organic compound is __________. (Nearest integer)
Given :
(a) The vapour pressure of water of $280 \mathrm{~K}$ is $14.2 \mathrm{~mm} \,\mathrm{Hg}$.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 27th July Evening Shift
In the above reaction, $5 \mathrm{~g}$ of toluene is converted into benzaldehyde with $92 \%$ yield. The amount of benzaldehyde produced is ______________ $\times 10^{-2} \mathrm{~g}$. (Nearest integer)
Correct Answer: 530
Explanation:
Moles $=\frac{5}{92}$
Moles of benzaldehyde produced $=\frac{5}{92} \times 0.92=0.05$
$\therefore$ Mass of benzaldehyde formed
$=0.05 \times 106$
$=5.3 \mathrm{~g}$
$=530 \times 10^{-2}$
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 25th July Morning Shift
While estimating the nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from $0.25 \mathrm{~g}$ of the compound neutralized $2.5 \mathrm{~mL}$ of $2 \,\mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}$. The percentage of nitrogen present in organic compound is ______________.
Correct Answer: 56
Explanation:
$\mathrm{NH}_{3}$ gas is neutralized by $2.5 \mathrm{~mL}$ of $2 \mathrm{M} \,\,\mathrm{H}_{2} \mathrm{SO}_{4}$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 30th June Morning Shift
An organic compound with 51.6% sulfur is heated in a Carius tube. The amount of this compound which will form 0.752 g of barium sulphate is ___________ $\times$ 10$-$1 g.
(Given molar mass of barium sulphate 233 g mol$-$1) (Nearest integer).
Sulfur $\%=\frac{\text { Weight of Sulphur }}{\text { Weight of Compound }} \times 100$
$51.6=\frac{0.1005}{\text { Weight of compound }} \times 100$
$\Rightarrow$ Weight of compound $=\frac{0.1005}{51.6} \times 100=1.9 \times 10^{-1} \mathrm{~g}$
$
\approx 2 \times 10^{-1} \mathrm{~g}
$
2022
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 29th June Morning Shift
Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia evolved from 0.55 g of the compound neutralised 12.5 mL of 1 M H2SO4 solution. The percentage of nitrogen in the compound is _____________. (Nearest integer)
Correct Answer: 64
Explanation:
Meq. of $\mathrm{H}_2 \mathrm{SO}_4=$ Meq. of $\mathrm{NH}_3$
12. $5 \times 1 \times 2=25$ meq. of $\mathrm{NH}_3$ $=25$ millimoles of $\mathrm{NH}_3$
So Millimoles of ${ }^{\prime} \mathrm{N}^{\prime}=25$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 28th June Morning Shift
A 2.0 g sample containing MnO2 is treated with HCl liberating Cl2. The Cl2 gas is passed into a solution of KI and 60.0 mL of 0.1 M Na2S2O3 is required to titrate the liberated iodine. The percentage of MnO2 in the sample is _____________. (Nearest integer)
[Atomic masses (in u) Mn = 55; Cl = 35.5; O = 16, I = 127, Na = 23, K = 39, S = 32]
Correct Answer: 13
Explanation:
First Step :
MnO2 + 4HCl $\to$ MnCl2 + Cl2 + 2H2O
Here 1 mol of MnO2 produce 1 mol of Cl2
$\therefore$ Mole ratio of ${n_{Mn{O_2}}}:{n_{C{l_2}}} = 1:1$
Second Step :
Cl2 + 2KI $\to$ 2KCl + I2
Here, 1 mol of Cl2 produce 1 mol of I2
Mole ratio of ${n_{C{l_2}}}:{n_{{I_2}}} = 1:1$
Third Step :
I2 + 2Na2S2O3 $\to$ 2NaI + Na2S2O3
1 mol of I2 react with 2 mol of Na2S2O3
Mole ratio of ${n_{{I_2}}}:{n_{{N_2}{S_2}{O_3}}} = 1:2$
Given Na2S2O3 is 60 mL of 0.1 M
$\therefore$ Number of moles of Na2S2O3
= V (in L) $\times$ M (Molarity)
= ${{60} \over {1000}} \times 0.1$
= 0.006 mol
$\therefore$ Number of moles of I2
$ = {1 \over 2}(0.006)$
= 0.003
$\therefore$ Moles of MnO2 = 0.003 (as mole ratio of MnO2 and Cl2 = 1 : 1)
Molar mass of MnO2 = 55 + 32 = 87
$\therefore$ Mass of MnO2 = 0.003 $\times$ 87 = 0.261 gm
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 28th June Morning Shift
In the estimation of bromine, 0.5 g of an organic compound gave 0.40 g of silver bromide. The percentage of bromine in the given compound is _________ % (nearest integer)
(Relative atomic masses of Ag and Br are 108u and 80u, respectively).
Correct Answer: 34
Explanation:
$188 \mathrm{~g} ~ \mathrm{AgBr}$ has $80 \mathrm{~g}$ of $\mathrm{Br}$
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 27th June Evening Shift
0.25 g of an organic compound containing chlorine gave 0.40 g of silver chloride in Carius estimation. The percentage of chlorine present in the compound is ___________. [in nearest integer]
(Given : Molar mass of Ag is 108 g mol$-$1 and that of Cl is 35.5 g mol$-$1)
Correct Answer: 40
Explanation:
Given, weight of organic compound = 0.25 g
Moles of AgCl = ${{0.4} \over M}$
Molecular mass of AgCl (M) = 143.5 gm
$\therefore$ Moles of AgCl = ${{0.4} \over 143.5}$
$\therefore$ Mass of Cl = ${{0.4} \over 143.5}$ $\times$ 35.5
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 24th June Evening Shift
0.2 g of an organic compound was subjected to estimation of nitrogen by Dumas method in which volume of N2 evolved (at STP) was found to be 22.400 mL. The percentage of nitrogen in the compound is _________. [nearest integer]
(Given : Molar mass of N2 is 28 g mol$-$1. Molar volume of N2 at STP : 22.4 L)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th August Evening Shift
Which one of the following tests used for the identification of functional groups in organic compounds does not use copper reagent?
A.
Barfoed's test
B.
Seliwanoff's test
C.
Benedict's test
D.
Biuret test for peptide bond
Correct Answer: B
Explanation:
In Seliwanoff's reagent, Cu is not present.
In Barfoed, Biuret and in Benediet reagent Cu is present.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th August Morning Shift
Acidic ferric chloride solution on treatment with excess of potassium ferrocyanide gives a Prussian blue coloured colloidal species. It is :
A.
Fe4[Fe(CN)6]3
B.
K5Fe[Fe(CN)6]2
C.
HFe[Fe(CN)6]
D.
KFe[Fe(CN)6]
Correct Answer: A
Explanation:
Prussian blue is a complex compound that has the formula $ \text{Fe}_{4}[\text{Fe(CN)}_{6}]_{3} $. It is formed by the reaction of ferric ions with ferrocyanide ions.
The reaction can be explained as follows:
Potassium ferrocyanide, $ \text{K}_{4}[\text{Fe(CN)}_{6}] $, dissociates in water to release ferrocyanide ions, $ [\text{Fe(CN)}_{6}]^{4-} $.
The ferric ions, $ \text{Fe}^{3+} $, present in the acidic ferric chloride solution will react with the ferrocyanide ions.
In an acidic medium, the ferric ions hydrolyze, forming a complex with ferrocyanide ions, leading to the formation of Prussian blue.
The Prussian blue complex is a colloidal species and exhibits an intense blue color. It has been historically used as a pigment in paints and also has applications in medical treatments as an antidote for certain types of heavy metal poisoning.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th August Morning Shift
Which one of the following complexes is violet in colour?
A.
[Fe(CN)6]4$-$
B.
[Fe(SCN)6]4$-$
C.
Fe4[Fe(CN6)]3 . H2O
D.
[Fe(CN)5NOS]4$-$
Correct Answer: D
Explanation:
(a) [Fe(CN)6]4$-$ $\to$ Pale yellow solution
(b) [Fe(SCN)6]4$-$ $\to$ Blood red colour
(c) Fe4[Fe(CN6)]3 . H2O $\to $ Prussian blue
(d) [Fe(CN)5NOS]4$-$ $\to$ Violet colour
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 20th July Evening Shift
In Carius method, halogen containing organic compound is heated with fuming nitric acid in the presence of :
A.
HNO3
B.
AgNO3
C.
CuSO4
D.
BaSO4
Correct Answer: B
Explanation:
Organic compound is heated with fuming nitric acid in the presence of silver nitrate in carius method.
Lunar caustic (AgNO3) is used as reagent hare to distinguish Cl$-$, Br- and I$-$ respectively as follows.
$C{l^ - }(aq)\buildrel {AgN{O_3}} \over
\longrightarrow AgCl{ \downarrow _{ppt}}$ white
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 20th July Morning Shift
An inorganic Compound 'X' on treatment with concentrated H2SO4 produces brown fumes and gives dark brown ring with FeSO4 in presence of concentrated H2SO4. Also Compound 'X' gives precipitate 'Y', when its solution in dilute HCl is treated with H2S gas. The precipitate 'Y' on treatment with concentrated HNO3 followed by excess of NH4OH further gives deep blue coloured solution, Compound 'X' is :
A.
Co(NO3)2
B.
Pb(NO2)2
C.
Cu(NO3)2
D.
Pb(NO3)2
Correct Answer: C
Explanation:
Compound ‘X’ is copper nitrate, i.e. Cu(NO3)2 is an inorganic
compound. On treatment with concentrated H2SO4 produces brown
fumes and gives dark brown ring with FeSO4
in presence of
concentrated H2SO4. Chemical reaction is as follows
Cu2+
is a group II cation with group II reagents
(HCl/H2S)
, it gives black coloured precipitates. These precipitates
gives blue colour solution on treatment with HNO3
followed by
excess of NH4OH.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 18th March Morning Shift
Reagent, 1-napthylamine and sulphanilic acid in acetic acid is used for the detection of
A.
N2O
B.
NO$_3^ - $
C.
NO
D.
NO$_2^ - $
Correct Answer: D
Explanation:
1-naphthyl amine and sulphanilic acid in acetic acid
is used for the detection of $NO_2^ - $.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 16th March Evening Shift
Match List - I with List - II :
List - I Test/Reagents/Observation(s)
List - II Species detected
(a)
Lassaigne's Test
(i)
Carbon
(b)
Cu(II) oxide
(ii)
Sulphur
(c)
Silver nitrate
(iii)
N, S, P, and halogen
(d)
The sodium fusion extract gives black precipitate with acetic acid and lead acetate
(iv)
Halogen Specifically
The correct match is :
A.
(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
B.
(a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
C.
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
D.
(a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
Correct Answer: C
Explanation:
(i) Lassaigne test is used to detect N, S, P, X element
(ii) Carbon and hydrogen are detected by heating the compound with copper (ii) oxide.
(iii) Halides are detected by silver nitrate
(iv) Sodium fusion extract gives black ppt. with acetic acid and lead acetate to confirm the presence of
sulphur.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th February Evening Shift
Seliwanoff test and Xanthoproteic test are used for the identification of ___________ and ________ respectively.
A.
ketoses, proteins
B.
proteins, ketoses
C.
aldoses, ketoses
D.
ketoses, aldoses
Correct Answer: A
Explanation:
Seliwanoff test and Xanthaproteic test are used for identification of 'Ketoses' and proteins
respectively.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th February Morning Shift
Which of the following is 'a' FALSE statement ?
A.
Carius method is used for the estimation of nitrogen in an organic compound.
B.
Kjeldahl's method is used for the estimation of nitrogen in an organic compound.
C.
Carius tube is used in the estimation of sulphur in an organic compound.
D.
Phosphoric acid produced on oxidation of phosphorus present in an organic compound is precipitated as Mg2P2O7 by adding magnesia mixture.
Correct Answer: A
Explanation:
Carius method is used in the estimation of
halogen and sulphur in organic compounds.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 25th February Evening Shift
Which of the following compound is added to the sodium extract before addition of silver nitrate for testing of halogens?
A.
Nitric acid
B.
Sodium hydroxide
C.
Ammonia
D.
Hydrochloric acid
Correct Answer: A
Explanation:
NaCN + HNO3 $ \to $ NaNO3 + HCN $ \uparrow $
Na2S + HNO3 $ \to $ NaNO3 + H2S $ \uparrow $
Nilnic acid decomposed NaCN & Na2S, else they precipitate in test & misquite the resolve
For testing of halogens, Nitric acid is added
to the sodium extract because if CN–
or S2–
are present then they will be oxidised and
removed before the test of halides.
2021
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 31st August Evening Shift
The transformation occurring in Duma's method is given below :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th August Morning Shift
The number of moles of CuO, that will be utilized in Dumas method for estimation nitrogen in a sample of 57.5 g of N, N-dimethylaminopentane is _____________ $\times$ 10$-$2. (Nearest integer)
Correct Answer: 1125
Explanation:
Moles of N in N, N - dimethylaminopentane = $\left( {{{57.5} \over {115}}} \right)$ = 0.5 mol
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th August Morning Shift
When 10 mL of an aqueous solution of KMnO4 was titrated in acidic medium, equal volume of 0.1 M of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The strength of KMnO4 in grams per litre is __________ $\times$ 10$-$2. (Nearest integer)
[Atomic mass of K = 39, Mn = 55, O = 16]
Correct Answer: 316
Explanation:
Let molarity of KMnO4 = x
(Equivalents of KMnO4 reacted) = (Equivalents of FeSO4 reacted)
$\Rightarrow$ (5 $\times$ x $\times$ 10 ml) = 1 $\times$ 0.1 $\times$ 10 ml
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th August Evening Shift
In the sulphur estimation, 0.471 g of an organic compound gave 1.44 g of barium sulphate. The percentage of sulphur in the compound is ____________%. (Nearest integer)
(Atomic Mass of Ba = 137 u)
Correct Answer: 42
Explanation:
Molecular mass of BaSO4 = 233 g
$\because$ 233 BaSO4 contain $\to$ 32 g sulphur
$\therefore$ 1.44 g BaSO4 contain $\to$ ${{32} \over {233}}$ $\times$ 1.44 g sulphur given : 0.471 g of organic compound
% of S = ${{32 \times 1.44} \over {233 \times 0.471}} \times 100 = 41.98\% \approx 42\% $
2021
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 27th July Morning Shift
An organic compound is subjected to chlorination to get compound A using 5.0 g of chlorine. When 0.5 g of compound A is reacted with AgNO3 [Carius Method], the percentage of chlorine in compound A is ____________ when it forms 0.3849 g of AgCl. (Round off to the Nearest Integer)
(Atomic masses of Ag and Cl are 107.87 and 35.5 respectively)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 25th July Evening Shift
0.8 g of an organic compound was analyzed by Kjeldahl's method for the estimation of nitrogen. If the percentage of nitrogen in the compound was found to be 42%, then ______________ mL of 1 M H2SO4 would have been neutralized by the ammonia evolved during the analysis.
Correct Answer: 12
Explanation:
Organic compound : 0.8 gm
wt. of N = $\left( {{{42} \over {100}} \times 0.8} \right)$ gm
mole of N = ${{{42 \times 0.8} \over {100 \times 14}} = {{2.4} \over {100}}}$ mol
moles of NH3 = ${{{2.4} \over {100}}}$
${{{1.2} \over {100}}}$ = 1 $\times$ V(l)
${ \Rightarrow {V_{{H_2}S{O_4}}} = {{1.2} \over {100}}l = 12}$ ml
2021
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 22th July Evening Shift
Methylation of 10 g of benzene gave 9.2 g of toluene. Calculate the percentage yield of toluene __________. (Nearest integer)
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 17th March Morning Shift
In the above reaction, 3.9 g of benzene on nitration gives 4.92 g of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is ________%. (Round off to the Nearest Integer).
(Given atomic mass : C : 12.0 u, H : 1.0 u, O : 16.0 u, N : 14.0 u)
Correct Answer: 80
Explanation:
Weight of (C6H5NO2)Theoretical = ${{{3.9} \over {78}} \times 123}$ = 6.15 g
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 16th March Evening Shift
In Duma's method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen collected at 287 K and 758 mm of Hg pressure. The percentage composition of nitrogen in the compound is __________. (Round off to the Nearest Integer). [Given : Aqueous tension at 287 K = 14 mm of Hg]
Correct Answer: 19
Explanation:
Aqueous tension at 287 K = 14 mm of Hg.
Hence actual pressure = (758 – 14)
= 744 mm of Hg.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 2nd September Evening Slot
If you spill a chemical toilet cleaning liquid on
your hand, your first aid would be
A.
aqueous NaOH
B.
aqueous NaHCO3
C.
aqueous NH3
D.
vinegar
Correct Answer: B
Explanation:
Toilet cleaning liquid has about 10.5% w/v HCl ; to neutralise its affect aqueous NaHCO3 is used while
NaOH is avoided for this purpose because of its highly corosive in nature and can burn body.
2020
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 2nd September Morning Slot
In Carius method of estimation of halogen,
0.172 g of an organic compound showed
presence of 0.08 g of bromine. Which of these
is the correct structure of the compound?
A.
H3C – CH2 – Br
B.
H3C – Br
C.
D.
Correct Answer: C
Explanation:
In Carius method
mass of organic compound = 0.172 gm
mass of Bromine = 0.08 gm
Hence % of Bromine = ${{0.08} \over {0.172}} \times 100$
The given table provides the results of three tests: Molisch's test, Barfoed's test, and Biuret test for three different biomolecules A, B, and C.
Molisch's Test: Detects carbohydrates. Positive results for A and B indicate that they are carbohydrates.
Barfoed's Test: Specifically for detecting monosaccharides. A positive result for B indicates that B is a monosaccharide, while A's negative result indicates it's not a monosaccharide, implying it might be a disaccharide.
Biuret Test: Used to detect the presence of proteins. Positive result for C indicates that it is a protein.
Given the options, we can analyze them as follows:
Option A: Lactose (disaccharide), Glucose (monosaccharide), Albumin (protein) - fits the given results.
Option B: Lactose (disaccharide), Fructose (monosaccharide), Alanine (amino acid) - not a match, as Alanine won't give a positive Biuret test.
Option C: Lactose (disaccharide), Glucose (monosaccharide), Alanine (amino acid) - same issue as Option B.
Option D: Glucose (monosaccharide), Fructose (monosaccharide), Albumin (protein) - not a match, as A must be a disaccharide.
Therefore, the correct option is:
Option A: A = Lactose, B = Glucose, C = Albumin.
2020
JEE Mains
Numerical
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2020 (Online) 6th September Morning Slot
In an estimation of bromine by Carius method,
1.6 g of an organic compound gave 1.88 g of
AgBr. The mass percentage of bromine in the
compound is _______.
(Atomic mass, Ag = 108, Br = 80 g mol–1)
Correct Answer: 50
Explanation:
Mass of organic compound = 1.6 gm
Mass of AgBr = 1.88 gm
Moles of Br = Moles of AgBr = ${{1.88} \over {188}}$ = 0.01
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 12th April Morning Slot
An organic compound 'A' is oxidizod with Na2O2 followed by boiling with HNO3. The resultant solution is
then treated with ammonium molybdate to yield a yellow precipitate.
Based on above observation, the
element present in the given compound is:
A.
Fluorine
B.
Phosphorus
C.
Nitrogen
D.
Sulphur
Correct Answer: B
Explanation:
Phosphorus is detected in the form of canary
yellow ppt on reaction with ammonium
molybdate.
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th April Morning Slot
The organic compound that gives following
qualitative analysis is :
Test
Inference
(a)
Dil. HCl
Insoluble
(b)
NaOH solution
soluble
(c)
Br2/water
Decolourization
A.
B.
C.
D.
Correct Answer: A
Explanation:
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 8th April Morning Slot
An organic compound neither reacts with neutral ferric chloride solution nor with fehling solution.
it however, reacts with Grignard reagent and gives positive iodoform test. The compound is :
A.
B.
C.
D.
Correct Answer: D
Explanation:
(1) The organic compound does not react with natural FeCl3 solution. It means phenolic group is absent in the organic compound.
(2) The organic compound does not give Fehling solution test. It means aldehyde group (- CHO) is absent in the organic compound.
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 8th April Morning Slot
An organic compound 'X' showing the following solubility profile is :
A.
o-Toluidine
B.
Benzamide
C.
Oleic acid
D.
m-Cresol
Correct Answer: D
Explanation:
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 10th January Evening Slot
The correct match between item 'I' and item 'II' is :
Lysine is an amino acid which gives nindydrin test.
Furfural has aldehyde group that is why it reacts with 1-napthol to give purple colouration.
Benzyl alcohol undergoes reaction with ceric ammonium nitrate to give pink colouration.
Styrene reacts with KMnO4 and gives benzoic acid and brown MnO2.
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th January Evening Slot
The tests performed on compound X and their inferences are :
Test
Inference
(a)
2, 4 - DNP test
Coloured precipitate
(b)
Idoform test
Yellow precipitate
(c)
Azo-dye test
No dye formation
Compound 'X' is :
A.
B.
C.
D.
Correct Answer: B
Explanation:
$ \to $ 2, 4 $-$ DNP test is given by aldehyde on ketone
$ \to $ Iodoform test is given by compound having
group.
2019
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2019 (Online) 9th January Morning Slot
The correct match between Item-I and Item-II is :
Item-I (drug)
Item-II (test)
A
Chloroxylenol
P
Carbylamine test
B
Norethindrone
Q
Sodium hydrogen carbonet test
C
Sulphapyridine
R
Ferric chloride test
D
Penicillin
S
Bayer's test
A.
A $ \to $ R; B $ \to $ P; C $ \to $ S; D $ \to $ Q
B.
A $ \to $ Q; B $ \to $ S; C $ \to $ P; D $ \to $ R
C.
A $ \to $ R; B $ \to $ S; C $ \to $ P; D $ \to $ Q
D.
A $ \to $ Q; B $ \to $ P; C $ \to $ S; D $ \to $ R
Correct Answer: C
Explanation:
(1) To react with FeCl3, with benzene ring there should be a $-$OH group attached.
(2) For carbylamine test there should be $-$NH2 group in a compound.
(3) NaHCO3 reacts when there is $-$COOH group in a compound.
(4) For Bayer's test there should be double bond or tripple bond between two carbon.
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 16th April Morning Slot
In a complexometric titration of metal ion with ligand
M(Metal ion) + L(Ligand) $ \to $ C(Complex) end point is estimated spectrophoto - metrically (through light absorption). If 'M' and 'C' do not absorb light and only 'L' absorbs, then the titration plot between absorbed light (A) versus volume of ligand 'L' (V) would look like :
A.
B.
C.
D.
Correct Answer: B
Explanation:
Ligand concentration is maximum when the reaction has not been initiated. At that point, absorption of light by ligands will be maximum. As the reaction starts, ligand concentration decreases so is the absorption of light till the point where all the metal ions are consumed and further complex formation does not take place.
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Online) 16th April Morning Slot
For standardizing NaOH solution, which of the following is used as a primary standard ?
A.
Ferrous Ammonium Sulfate
B.
dil. HCl
C.
Oxalic acid
D.
Sodium tetraborate
Correct Answer: C
Explanation:
The solution that can be prepared directly by dissolving an accurately weighed amount of the substance in water and making up the volume to a known amount by water is known as primary standard solution. Oxalic acid is an example of primary standard while NaOH is a secondary standard solution. Hence, strength of NaOH solution can be determined by titrating against oxalic acid. This process is known as standardisation.
2018
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2018 (Offline)
Which of the following compounds will be suitable for Kjeldahl’s method for nitrogen estimation?
A.
B.
C.
D.
Correct Answer: C
Explanation:
In Kjeldahl’s method, ammonium ion or ammonia should generate from a compound.
So, those compounds which contain nitrogen in nitro and azo groups and nitrogen in the ring cannot generate ammonium ion or ammonia.
Among the given option only aniline can generate NH3. So it is suitable far nitrogen estimation by Kjeldahl’s method.
2017
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2017 (Offline)
Sodium salt of an organic acid ‘X’ produces effervescence with conc. H2SO4. ‘X’ reacts with the acidified
aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of KMnO4. ‘X’ is :
A.
CH3COONa
B.
Na2C2O4
C.
C6H5COONa
D.
HCOONa
Correct Answer: B
Explanation:
2016
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2016 (Online) 10th April Morning Slot
Sodium extract is heated with concentrated HNO3 before testing for halogens because :
A.
Silver halides are totally insoluble in nitric acid.
B.
Ag2S and AgCN are soluble in acidic medium.
C.
S2− and CN−, if present, are
decomposed by conc. HNO3 and hence do not interfere in the test.
D.
Ag reacts faster with halides in acidic medium.
Correct Answer: C
Explanation:
Sodium fusion extract is heated with dilute HNO3 to decompose NaCN or Na2S (if present) to HCN and H2S gases, respectively.
The presence of N or S interferes with the test of halogens by forming precipitate with AgNO3.
29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s method
and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid
required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in
the compound is
A.
59.0
B.
47.4
C.
23.7
D.
29.5
Correct Answer: C
Explanation:
Moles of $HCl$ taken $ = 20 \times 0.1 \times {10^{ - 3}} = 2 \times {10^{ - 3}}$
