datavorous/entrance-exam-dataset huggingface dataset:
In chromyl chloride, the number of d-electrons present on chromium is same as in (Given at no. of Ti : 22, V : 23, Cr : 24, Mn : 25, Fe : 26)

Options:
A V (IV)
B Mn (VII)
C Fe (III)
D Ti (III)

The correct answer is:
Mn (VII)

Solution:
In chromyl chloride (CrO2Cl2), the oxidation state of chromium (Cr) is +6.

So, Cr (VI) = [Ar]18 3d0

For the given elements the configuration is,Mn(VII) = [Ar]18 3d0

Fe(III) = [Ar]18 3d5

Ti(III) = [Ar]18 3d1

V(IV) = [Ar]18 3d1

So among the given options Mn(VII) and Cr (VI) will have same number of d-electrons.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (08 Apr Shift 1)
In Chromyl chloride, the oxidation state of chromium is + ___________.

Options:

The correct answer is:
6

Solution:
The chemical formula of Chromyl chloride is CrO2Cl2. The oxidation state of chromium in chromyl chloride (CrO2Cl2) is +6.

Let the oxidation number of Chromium be “x”, the oxidation state of oxygen is -2, the oxidation state of chlorine is -1.

Then,

x + 2(-2) + 2(-1) = 0

x-6=0x=6

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (15 Apr Shift 1)
In Clemmensen Reduction carbonyl compound is treated with

Options:
A Zinc amalgam \( +\mathrm{HCl} \)
B Sodium amalgam \( +\mathrm{HCl} \)
C Zinc amalgam + nitric acid
D Sodium amalgam \( +\mathrm{HNO}_{3} \)

The correct answer is:
Zinc amalgam \( +\mathrm{HCl} \)

Solution:
Clemmensen reduction is a chemical reaction in which reduction of ketones (or aldehydes) to alkanes takes place using zinc amalgam and concentrated hydrochloric acid. The original Clemmensen reduction conditions are particularly effective at reducing aryl-alkyl ketones, such as those formed in a Friedel-Craft's acylation.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In Clemmensen's reduction, the catalyst used is

Options:
A $\mathrm{Zn}-\mathrm{Hg}+$ Conc $\mathrm{HCl}$
B $\mathrm{NH}_{2} \mathrm{NH}_{2}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONa}$
C $\mathrm{PdCl}_{2} / \mathrm{H}_{2} \mathrm{O}$
D $\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3} \mathrm{P}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONa}$

The correct answer is:
$\mathrm{Zn}-\mathrm{Hg}+$ Conc $\mathrm{HCl}$

Solution:
Zinc amalgam $(\mathrm{Zn}-\mathrm{Hg})$ and conc $\mathrm{HCl}$ is used as a catalyst for Clemmensen's reduction. This reagent converts $\sum \mathrm{C}=\mathrm{O}$ group to
$>\mathrm{CH}_{2}$
group.

Topic: Chemistry, Hydrocarbons, JEE Main
In cockroach, larval and nymphal characters are maintained by

Options:
A ecdysone
B salivary glands
C parotid gland
D juvenile hormone

The correct answer is:
juvenile hormone

Solution:
Juvenile hormone is produced by corpora allata in insects, it favours the development of juvenile characteristics. During larval life, this hormone predominates and each moult yields another larger juvenile and keeps the larva in immature condition or maintains juvenility.Ecdysone is a steroidal, moulting hormone of the prothoracic glands, named "ecdysone" in insects that triggers metamorphosis.Salivary glands are the merocrine or epicrine type of gland. Their secretion diffuses out through the cell surface, causing no damage to the cell. In man, 3 pairs of salivary glands are present and in rabbit, they are 4 pairs. Salivary glands are absent in frog.Parotid glands are compound tubulo-acinar glands and are the type of salivary glands with a flask-like secretory portion. Poison gland of a toad is modified parotid gland. In rabbit, parotid glands are the largest salivary gland, present beneath the ears. Duct of parotid gland is known as Stenson's duct in man.

Topic: Biology, Structural Organisation in Animals, JIPMER, JIPMER 2009
In comparison to boron, berylium has:

Options:
A greater nuclear charge and lesser first ionisation enthalpy
B lesser nuclear charge and greater first ionisation enthalpy
C lesser nuclear charge and lesser first ionisation enthalpy.
D greater nuclear charge and greater first ionisation enthalpy.

The correct answer is:
lesser nuclear charge and greater first ionisation enthalpy

Solution:
Electronic configuration of Be =1s2 2s2
Electronic configuration of B =1s2 2s2 2p1
Due to full-filled configuration of Beryllium, it has higher first ionization energy.
Effective nuclear charge
Zeff of Be =4-σ=4-2×0.85+0.35×1=2.05
Zeff of B =5-σ=5-2×0.85+0.35×2=2.6
∴Zeff of B > Zeff of Be

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2019 (12 Apr Shift 2)
In comparison to the zeolite process for the removal of permanent hardness, the synthetic resin method is

Options:
A less efficient as it exchanges only anions.
B more efficient as it can exchange both cations as well as anions.
C less efficient as the resins cannot be regenerated.
D more efficient as it can exchange only cations.

The correct answer is:
more efficient as it can exchange both cations as well as anions.

Solution:
Zeolite removes only cations from hard water, as given below by reaction:

2 NaZ + M+2(aq)→MZ2 + 2 Na+ Here M-Ca, Mg

While, synthetic resin removes both cations and anions, as given by following reactions:

For cations-

2 RNa(s) +M+2(aq)→ R2M (s)+ 2Na+(aq)

For anions-

RNH3+OH-(s) + X-(aq) →RNH3+X- (s)+ OH-(aq)

So, synthetic resin method is more efficient as it can exchange both cations as well anions.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2020 (07 Jan Shift 1)
In compound C, H and N atoms are present in 9 : 1 : 3.5 by weight. Molecular weight of compound is 108. Molecular formula of compound is

Options:
A C2H6N2
B C3H4N
C C6H8N2
D C9H12N3

The correct answer is:
C6H8N2

Solution:
Percentage	R.N.A.	Simplest ratio
C	9	912=34	3
H	1	11=1	4
N	35	3.514=14	1

Empirical formula =C3H4N
C3H4Nn=108,12×3+4×1+14n=108
54n=108
n=10854=2
So molecular formula =2×C3H4N=C6H8N2

Topic: Chemistry, Hydrocarbons, JEE Main
In compound O2SCNH22, the geometry around S and N are respectively -

Options:
A trigonal planar, trigonal pyramidal
B tetrahedral, pyramidal
C trigonal planar, tetrahedral
D linear, pyramidal

The correct answer is:
trigonal planar, trigonal pyramidal

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In connection with the circuit drawn below, the value of current flowing through 2 kΩ resistor is______×10-4 A.

Options:

The correct answer is:
25

Solution:
Current through 2kΩ resistance

I=52×103=2.5×10-3 A

I=25×10-4 A

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (24 Feb Shift 1)
In context of the lanthanoids, which of the following statements is not correct ?

Options:
A
All the members exhibit $+3$ oxidation state

B
Because of similar properties the separation of lanthanoids is not easy.

C
Availability of $4 \mathrm{f}$ electrons results in the formation of compounds in $+4$ state for all the members of the series.

D
There is a gradual decrease in the radii of the members with increasing atomic number in the series.

The correct answer is:
Availability of $4 \mathrm{f}$ electrons results in the formation of compounds in $+4$ state for all the members of the series.

Solution:
The general o.s of lanthanides is $+3$, only few elements exhibit $+4$ o.s.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2011
In context with the industrial preparation of hydrogen from water gas $\left(\mathrm{CO}+\mathrm{H}_2\right)$, which of the following is the correct statement?

Options:
A
$\mathrm{CO}$ and $\mathrm{H}_2$ are fractionally separated using differences in their densities

B
$\mathrm{CO}$ is removed by absorption in aqueous $\mathrm{Cu}_2 \mathrm{Cl}_2$ solution

C
$\mathrm{H}_2$ is removed through occlusion with $\mathrm{Pd}$

D
$\mathrm{CO}$ is oxidised to $\mathrm{CO}_2$ with steam in the presence of a catalyst followed by absorption of $\mathrm{CO}_2$ in alkali

The correct answer is:
$\mathrm{CO}$ is oxidised to $\mathrm{CO}_2$ with steam in the presence of a catalyst followed by absorption of $\mathrm{CO}_2$ in alkali

Solution:

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2008
In context with the industrial preparation of hydrogen from water gas \( \left(\mathrm{CO}+\mathrm{H}_{2}\right) \), which of the following is the correct statement?

Options:
A \( \mathrm{CO} \) is oxidised to \( \mathrm{CO}_{2} \) with steam, in the presence of a catalyst followed by the dissolution of \( \mathrm{CO}_{2} \) in water under
ssure.
B \( \mathrm{CO} \) and \( \mathrm{H}_{2} \) are fractionally separated using the differences in their densities.
C \( \mathrm{CO} \) is removed by absorption in aqueous \( \mathrm{Cu}_{2} \mathrm{Cl}_{2} \) solution.
D \( \mathrm{H}_{2} \) is removed through occlusion with \( \mathrm{Pd} \).

The correct answer is:
\( \mathrm{CO} \) is oxidised to \( \mathrm{CO}_{2} \) with steam, in the presence of a catalyst followed by the dissolution of \( \mathrm{CO}_{2} \) in water under
ssure.

Solution:
CO is oxidised to CO2 with steam in the presence of a catalyst, followed by the dissolution of CO2 in water under pressure.

CO+H2water gas+H2Osteam→catlaystCO2+2H2

And,

CO & H2 are not fractionally separated by using the differences in their densities. Carbon monoxide is separated from hydrogen gas by passing it through water at a pressure of 30 atmospheres due to which carbon monoxide is absorbed in water unlike hydrogen gas

CO is not removed by absorption in aqueous Cu2Cl2 solution.

H2 is not removed through occlusion with Pd

Topic: Chemistry, Hydrogen, JEE Main
In context with the transition elements, which of the following statements is incorrect?

Options:
A
In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes.

B
In the highest oxidation states, the transition metal show basic character and form cationic complexes.

C
In the highest oxidation states of the first five transition elements (Sc to Mn), all the $4 s$ and $3 d$ electrons are used for bonding.

D
Once the $d^5$ configuration is exceeded, the tendency to involve all the 3d electrons in bonding decreases.

The correct answer is:
In the highest oxidation states, the transition metal show basic character and form cationic complexes.

Solution:
In higher Oxidation states transition elements show acidic nature

Topic: Chemistry, d and f Block Elements, JEE Main
In context with the transition elements, which of the following statements is incorrect?

Options:
A
In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes.

B
In the highest oxidation states, the transition metal show basic character and form cationic complexes.

C
In the highest oxidation states of the first five transition elements (Sc to Mn), all the $4 s$ and $3 d$ electrons are used for bonding.

D
Once the $d^5$ configuration is exceeded, the tendency to involve all the 3d electrons in bonding decreases.

The correct answer is:
In the highest oxidation states, the transition metal show basic character and form cationic complexes.

Solution:
In higher Oxidation states transition elements show acidic nature

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2009
In conversion of lime-stone to lime, $\mathrm{CaCO}_3(\mathrm{~s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$
the vales of $\Delta \mathrm{H}^{\circ}$ and $\Delta \mathrm{S}^{\circ}$ are $+179.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $160.2 \mathrm{~J} / \mathrm{K}$ respectively at $298 \mathrm{~K}$ and $1\mathrm{~bar}$.
Assuming that $\Delta \mathrm{H}^{\circ}$ do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is

Options:
A
$1008 \mathrm{~K}$

B
$1200$

C
$845 \mathrm{~K}$

D
$1118 \mathrm{~K}$

The correct answer is:
$1118 \mathrm{~K}$

Solution:
We know, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$
So, lets find the equilibrium temperature, i.e. at which
$\begin{aligned}
& \Delta \mathrm{G}=0 \\
& \Delta \mathrm{H}=\mathrm{T} \Delta \mathrm{S} \\
& \mathrm{T}=\frac{179.1 \times 1000}{160.2} \\
& =1118 \mathrm{~K}
\end{aligned}$
So, at temperature above this, the reaction will become spontaneous. Hence, (D) is correct answer.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2007
In coordination compound $\left[\mathrm{Co}(e n)_2 \mathrm{Cl}_2\right] \mathrm{Cl}$ which is false

Options:
A Show geometrical Isomerism
B Show optical Isomerism
C Show ionic Isomerism
D A octahedral complex

The correct answer is:
Show ionic Isomerism

Solution:
When coordinate compounds gives different ions in solution then it produces ionic isomerism while this situation is not present in $\left[\mathrm{Co}(e n)_2 \mathrm{Cl}_2\right] \mathrm{Cl}$.

Topic: Chemistry, Coordination Compounds, JEE Main
In curing cement plasters water is sprinkled from time to time. This helps in

Options:
A
developing interlocking needle-like crystals of hydrated silicates

B
hydrating sand and gravel mixed with cement

C
converting sand into silicic acid

D
keeping it cool

The correct answer is:
developing interlocking needle-like crystals of hydrated silicates

Solution:
No Solution Available

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2003
In curing cement plasters water is sprinkled from time to time. This helps in

Options:
A keeping it cool
B developing interlocking needle-like crystals of hydrated silicates
C hydrating sand and gravel mixed with cement
D converting sand into silicic acid

The correct answer is:
developing interlocking needle-like crystals of hydrated silicates

Solution:
Water develops interlocking needle-like crystals of hydrated silicates. The reactions involved are the hydration of calcium aluminates and calcium silicates which change into their colloidal gels. At the same time, some calcium hydroxide and aluminium hydroxides are formed as precipitates due to hydrolysis. Calcium hydroxide binds the particles of calcium silicates together while aluminium hydroxide fills the interstices rendering the mass impervious.

Topic: Chemistry, s Block Elements, JEE Main
In cyclic structure of SO3x if x is number of SO3 molecules involved in cycle formation and y is the number of sp3 hybridised S-atoms. What is the sum of x + y?

Options:

The correct answer is:
6.00

Solution:
x = 3, y = 3
So x + y = 6

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In cymose inflorescence

Options:
A Main axis terminates in a flower
B Main axis is of indefinite growth
C Flower are borne in basipetal order
D Both [a] and [c]

The correct answer is:
Both [a] and [c]

Solution:
In cymose main axis is of definite growth and terminates in flower. Flowers are produced laterally and are arranged in basipetal order in which old flowers are towards the apex and young flowers are towards the base.

Topic: Biology, Morphology of Flowering Plants, NEET
In $\mathrm{D}$ - glucose $(\mathrm{X})$ and $\mathrm{D}$ - Fructose $(\mathrm{Y})$, which hydroxyl group participates in hemiacetal formation

Options:
A $\mathrm{C}-5$ in both $\mathrm{X}$ and $\mathrm{Y}$
B $\mathrm{C}-5$ in $\mathrm{X}$ and $\mathrm{C}-6$ in $\mathrm{Y}$
C $C-6$ in $X$ and $C-5$ in $Y$
D $\mathrm{C}-4$ in both $\mathrm{X}$ and $\mathrm{Y}$

The correct answer is:
$\mathrm{C}-5$ in both $\mathrm{X}$ and $\mathrm{Y}$

Solution:
In both, glucose and fructose, C-5 atom participates in hemiacetal formation.

Topic: Chemistry, Biomolecules, JEE Main
In DCT, antidiuretic hormone makes the filtrate __________ to blood plasma.

Options:
A hypertonic
B hypotonic
C isotonic
D both A and B

The correct answer is:
hypertonic

Solution:
Nephron is the structural and functional part of the kidney. It consists of the renal corpuscle, and renal tubule. PCT is the main site of absorption as all essential nutrients and 70-80% of electrolytes and water are reabsorbed here. thus the filtrate remains isotonic here. The ascending limb of Henle's loop is impermeable to water and permeable to electrolytes.The filtrate from here passes into the DCT hence it is hypotonic to it.ADH hormone absorbs water from DCT and thus filtrate becomes hypertonic.

Topic: Biology, Excretory Products and their Elimination, NEET
In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is include? Sketch such a trajectory and explain why you have drawn it that way.

Options:

None

Solution:
When air resistance acts on projectile then its vertical and horizontal then particle energy as well as horizontal component of velocity keep on decreasing making the fall steeper than rise as shown in the figure.
When we are neglecting air resistance path was symmetric parabola $(O A B)$. When air resistance is considered path is asymmetric parabola $(\bar{O} A \bar{C})$.
By formula $R=\frac{u^2}{g} \sin 2 \theta$, and $H_{\max }=\frac{u^2 \sin ^2 2 \theta}{2 g}$
So, $\left(h_1 < h_2\right)$ and $\left(R_1 < R_2\right)$.
But time of flight for both will remain same.

Topic: Physics, Motion In Two Dimensions
In \( \Delta A B C, A D \) is perpendicular to \( B C \) and \( O \) is circumcentre of \( \Delta A B C \cdot \angle B=55^{\circ}, \angle A=80^{\circ} . \) Find \( \angle D A O \)

Options:
A \( 5^{\circ} \)
B \( 10^{\circ} \)
C \( 15^{\circ} \)
D \( 20^{\circ} \)

The correct answer is:
\( 10^{\circ} \)

Solution:
∠DAO=∠DAC-∠OAC

=90°-C-90°-B

=B-C

Now, by angle sum property of a triangle

A+B+C=180°

⇒C=180°-55°+80°=45°

Therefore,

⇒B-C=55°-45°=10°

Topic: Mathematics, Straight Lines, JEE Main
In deriving Bernoulli's equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy. (a) What is the largest average velocity of blood flow in an artery of diameter $2 \times 10^{-3}$ $m$ if the flow must remain laminar? (b) Do the dissipative forces become more important as the fluid velocity increase? Discuss qualitively.

Options:

None

Solution:
(a) If dissipative forces are present, then some forces in liquid flow due to pressure difference is spent against dissipative forces. Due to which the pressure drop becomes large.
(b) The dissipative forces become more important with increasing flow velocity, because of turbulence.

Topic: Physics, Mechanical Properties of Fluids
In deriving the kinetic gas equation, use is made of the root mean square velocity of the molecules because it is

Options:
A The average velocity of the molecules
B The most probable velocity of the molecules
C The square root of the average square velocity of the molecules
D The most accurate form in which velocity can be used in these calculations

The correct answer is:
The square root of the average square velocity of the molecules

Solution:
The square root of the average square velocity of the molecules

Topic: Chemistry, States of Matter, JEE Main
In diborane, the two $H-B-H$ angles are nearly

Options:
A $60^{\circ}, 120^{\circ}$
B $95^{\circ}, 120^{\circ}$
C $95^{\circ}, 150^{\circ}$
D $120^{\circ}, 180^{\circ}$

The correct answer is:
$95^{\circ}, 120^{\circ}$

Solution:
Dilthey in 1921 proposed a bridge structure for diborane. Four hydrogen atoms, two on the left and two on the right,known as terminal hydrogens and two boron atoms lie in the same plane.Two hydrogen atoms forming bridges,one above and other below, lie in a plane perpendicular to the rest of molecule.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In dilute aqueous H2SO4 , the complex diaquodioxalatoferrate(II) is oxidized by MnO4-
For this reaction, the ratio of the rate of change of H+ to the rate of change of MnO4- is

Options:

The correct answer is:
8

Solution:
The balanced chemical equation for the oxidation of the complex is as given below
MnO4-+5[FeH2OC2O4)22-+8H+→5FeH2OC2O42-+Mn2++4H2O
Hence -ddtMnO4-=-18ddtH+
ddtH+ddtMnO4-=8
Rate of change of H+ is 8 times the rate of change of MnO4- .

Topic: Chemistry, d and f Block Elements, JEE Main
In dispersion without deviation

Options:
A The emergent rays of all the colours are parallel to the incident ray
B Yellow coloured ray is parallel to the incident ray
C Only red coloured ray is parallel to the incident ray
D All the rays are parallel, but not parallel to the incident ray

The correct answer is:
Yellow coloured ray is parallel to the incident ray

Solution:
Deviation is zero only for a particular colour, it is generally taken to be yellow.

Topic: Physics, Ray Optics, JEE Main
In DNA, the complimentary bases are

Options:
A Uracil and adenine; cytosine and guanine
B Adenine and thymine; guanine and cytosine
C Adenine and thymine; guanine and uracil
D Adenine and guanine; thymine and cytosine

The correct answer is:
Adenine and thymine; guanine and cytosine

Solution:
$$
\begin{gathered}
A===T \\
G \equiv \equiv \equiv C
\end{gathered}
$$

Topic: Chemistry, Biomolecules, NEET, NEET 2008 (Mains)
In DNA, which is absent?

Options:
A Adenine
B Thymine
C Guanine
D Uracil

The correct answer is:
Uracil

Solution:
Uracil remains absent in DNA. It is present in RNA. The nitrogen bases of DNA and RNA are as follows :

Topic: Biology, Biomolecules (B), JIPMER, JIPMER 2006
In Dow's process, the starting raw material is

Options:
A phenol
B chlorobenzene
C aniline
D diazobenzene

The correct answer is:
chlorobenzene

Solution:
In Dow's process, firstly sodium phenoxide is obtained by the reaction of chlorobenzene with $\mathrm{NaOH}$ at $573-623 \mathrm{~K}$ and $200 \mathrm{~atm}$, which on acid hydrolysis forms phenol.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
In dry cell the reaction which takes place at the zinc anode is

Options:
A $\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}(\mathrm{s})$
B $Z n(s) \rightarrow Z n^{2+}+2 e$
C $\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}(\mathrm{s})$
D $M n(s) \rightarrow \mathrm{Mn}^{+}+\mathrm{e}^{-}+1.5 \mathrm{~V}$

The correct answer is:
$Z n(s) \rightarrow Z n^{2+}+2 e$

Solution:
At anode: $\mathrm{Zn} n_{(s)} \rightarrow \mathrm{Zn}^{2+}+2 e^{-}$

Topic: Chemistry, Electrochemistry, JEE Main
In Dumas method, $0.3 \mathrm{~g}$ of an organic compound gave $45 \mathrm{~mL}$ of nitrogen at STP. The percentage of nitrogen is

Options:
A $16.9$
B $18.7$
C $23.2$
D $29.6$

The correct answer is:
$18.7$

Solution:
In Dumes method,
$\%$ of nitrogen $=\frac{28 \mathrm{~V} \times 100}{22400 \times W}=\frac{28 \times 45 \times 100}{22400 \times 0.3}=18.75$

Topic: Chemistry, General Organic Chemistry, JEE Main
In Dumas method for the estimation of N2, the sample is heated with copper oxide and the gas evolved is passed over:

Options:
A Copper gauze
B Pd
C Ni
D Copper oxide

The correct answer is:
Copper gauze

Solution:
In Dumas method, the nitrogen-containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water.

Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide

So, the correct answer is Copper gauze.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (31 Jan Shift 2)
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]

Options:

The correct answer is:
19

Solution:
In Duma's method of estimation of Nitrogen. 0.1840 gm of organic compound gave 30 mL of nitrogen which is collected at 287 K and 758 mm of Hg.

Given ;

Aqueous tension at 287 K=14 mmof Hg.

Hence, actual pressure =758-14

=744 mm of Hg.

Volume of nitrogen at STP=273×744×30287×760

V=27.935 mL

∵ 22400 mL of N2 at STP weighs =28 gm.

∴ 27.94 mL of N2 at STP weighs =2822400×27.94 gm

=0.0349 gm

Hence % of Nitrogen =0.03490.1840×100

=18.97%

Round off. Answer =19%

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (16 Mar Shift 2)