datavorous/entrance-exam-dataset huggingface dataset:
Select the option with a correct representation of an animal belonging to the below mentioned class of vertebrates.

I. The notochord is persistent throughout life. The mouth is ventral.
II. The skin is moist without scales. The tail may be present.
III. 6 to 15 pairs of gill slits. Jaws absent.
IV. Viviparous usually. Pinna present in most of the members.

Options:
A

I	II	III	IV
Dog fish	Ichthyophis	Calotes	Canis

B

I	II	III	IV
Hagfish	Salamandra	Betta	Aptenodytes

C

I	II	III	IV
Pristis	Hyla	Lamprey	Pteropus

D

I	II	III	IV
Clarius	Rana	Myxine	Delphinus

The correct answer is:

I	II	III	IV
Pristis	Hyla	Lamprey	Pteropus

Solution:
I - Cartilaginous fish (Pristis - Saw fish), II - Amphibians (Salamandra - Salamandar), III - Cyclostomates (Lamprey - Petromyzon) and IV - Mammals (Pteropus - Flying fox - a bat).

Topic: Biology, Animal Kingdom, NEET
Select the pair of hormones having a similar mechanism/mode of action.

Options:
A Glucagon and cortisol
B Estrogen and LH
C Oxytocin and FSH
D Thyroxine and insulin

The correct answer is:
Oxytocin and FSH

Solution:
On the basis of their chemical nature, hormones can be divided into groups:
(i) Peptide, polypeptide, protein hormones (e.g., insulin, glucagon, pituitary hormones, hypothalamic hormones, etc.)
(ii) Steroids (e.g., cortisol, testosterone, estradiol and progesterone)
(iii) Iodothyronines (thyroid hormones)
(iv) Amino-acid derivatives (e.g., epinephrine).
Hormones which interact with membrane-bound receptors normally do not enter the target cell, but generate secondary messengers (e.g., cyclic AMP, IP3, Ca+2, etc.) which in turn regulate cellular metabolism.
Hormones which interact with intracellular receptors (e.g., steroid hormones, iodothyronines, etc.) mostly regulate gene expression or chromosome function by the interaction of hormone-receptor complex with the genome. Cumulative biochemical actions result in physiological and developmental effects.

Topic: Biology, Chemical Coordination and Integration, NEET
Select the set of incorrect statements from the ones given below.

A. Plasma is straw coloured and viscous in nature.
B. Eosinophils are 2-3% of total WBCs.
C. Lymphocytes are the second-most abundant agranulocytes.
D. Stroke volume is the amount of blood that each ventricle pumps during one minute.

Options:
A A, B and D
B C and D
C A and B
D A and D

The correct answer is:
C and D

Solution:
Lymphocytes are the second-most abundant (20-25%) white blood cells (WBCs) and the most abundant agranulocytes in the blood. The most abundant WBCs are the neutrophils (60-65%), which are one of the granulocytes.
During a cardiac cycle, each ventricle pumps out approximately 70 mL of blood. This is called the stroke or beat volume. Each cardiac cycle has a duration of 0.8 seconds.

Topic: Biology, Body Fluids and Circulation, NEET
Select the statement that is incorrect for cortical nephron.

Options:
A They are the most abundant of all the nephrons in kidneys
B Their glomeruli lie close to the inner margin of the cortex near medulla
C They control plasma volume when water supply is normal
D Loop of Henle is short in size

The correct answer is:
Their glomeruli lie close to the inner margin of the cortex near medulla

Solution:
Cortical nephrons are found in the cortex region of kidney. They have short Henle's loop which extend very little into the medulla section. Other type of nephron is called as juxta medullary nephron which are larger in size and have long loop extending deep into the medulla. The glomerulus of these nephron are placed close to the inner margin of the cortex.

Topic: Biology, Excretory Products and their Elimination, NEET
Select the wrong statement from the following.

Options:
A Ribozymes are nucleic acids with catalytic power.
B Nucleic acids serve as genetic material.
C Proteins, nucleic acids and polysaccharides are the only three types of macromolecules found in the living system.
D Collagen is the most abundant protein in the whole of the biosphere and RuBisCo is the most abundant protein in the animal world.

The correct answer is:
Collagen is the most abundant protein in the whole of the biosphere and RuBisCo is the most abundant protein in the animal world.

Solution:
-Collagen is the most abundant protein in the animal world and Ribulose bisphosphate carboxylase-oxygenase (RUBISCO) is the most abundant protein in the whole of the biosphere.
-Collagen is the most abundant protein in the human body
-Rubisco found in the stroma of chloroplast and has Bifunctional activity (carboxylase and oxygenase)
-Low CO2 high O2 → Rubisco perform oxygenase activity.
-High CO2 low O2 → Rubisco perform carboxylase activity

Topic: Chemistry, Biomolecules, NEET
Sensory organs like antennae, eyes (compound and simple), statocysts or balancing organs are present in members of phylum

Options:
A Mollusca
B Coelenterata
C Arthropoda
D Echinodermata

The correct answer is:
Arthropoda

Solution:
The body consists of a head, thorax, and abdomen. They have jointed appendages (arthros-joint, poda-appendages). Respiratory organs are gills, book gills, book lungs, or tracheal systems. The circulatory system is of open type. Sensory organs like antennae, eyes (compound and simple), statocysts, or balancing organs are present. Excretion takes place through malpighian tubules.

Topic: Biology, Animal Kingdom, NEET
Sets $A$ and $B$ have $n$ elements in common. How many elements will $(A \times B)$ and $(B \times A)$ have in common?

Options:
A 0
B 1
C n
D $n^{2}$

The correct answer is:
$n^{2}$

Solution:
The total number of elements common in $(A \times B)$ and $(B \times A)$ is $n^{2}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
Seven unbiased coins are tossed 128 times. In how many

Options:
A 99
B 102
C 103
D 104

The correct answer is:
99

Solution:
Let $\mathrm{X}$ denote the no. of coins showing 3 or more heads in a set of 7 coins. $\mathrm{X}$ follows binomial distribution with $\mathrm{n}=7$ $\mathrm{p}=$ probability of getting a head in a single toss of a coin
$\Rightarrow \mathrm{p}=\frac{1}{2} ;$ thus $\mathrm{q}=1-\mathrm{p}=\frac{1}{2}$
$\therefore$ Probability of getting at least 3 heads $=\mathrm{P}(\mathrm{x} \geq 3)$
$=1-\mathrm{P}(\mathrm{x} < 3)$
$=1-[\mathrm{P}(\mathrm{x}=0)+\mathrm{P}(\mathrm{x}=1)+\mathrm{P}(\mathrm{x}=2)]$
$=1-\left[{ }^{7} \mathrm{C}_{0}+{ }^{7} \mathrm{C}_{1}+{ }^{7} \mathrm{C}_{2}\right] \frac{1}{2^{7}}$
$=\frac{128}{128}-\frac{29}{128}=\frac{99}{128}$
$\therefore$ No. of throws $=\frac{99}{128} \times 128=99$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 2)
ùõÉ- sheets structure is which type of protein?

Options:
A Primary
B Secondary
C Tertiary
D Quaternary

The correct answer is:
Secondary

Solution:
Correct Option is : 2
Secondary

Topic: Chemistry, Biomolecules, NEET
Show that the motion of a particle represented by $y=\sin \omega t-\cos \omega t$ is simple harmonic with a period of $2 \pi / \omega$.

Options:

None

Solution:
When we convert the given combination of two simple harmonic functions to a single harmonic (sine or cosine) function it can be written uniquely in the form of $\cos \left(\frac{2 \pi}{T} t+\phi\right)$ or $\sin \left(\frac{2 \pi}{T} t+\phi\right)$
As given that the displacement function
$$
\begin{aligned}
y &=\sin \omega t-\cos \omega t \\
&=\sqrt{2}\left(\frac{1}{\sqrt{2}} \cdot \sin \omega t-\frac{1}{\sqrt{2}} \cdot \cos \omega t\right)
\end{aligned}
$$
$$
\begin{aligned}
&=\sqrt{2}\left[\cos \left(\frac{\pi}{4}\right) \cdot \sin \omega t-\sin \left(\frac{\pi}{4}\right) \cdot \cos \omega t\right] \\
&=\sqrt{2}\left[\sin \left(\omega t-\frac{\pi}{4}\right)\right]=\sqrt{2}\left[\sin \left(\omega t-\frac{\pi}{4}\right)\right]
\end{aligned}
$$
Comparing with standard SHM equation
$$
y=a \sin \left(\frac{2 \pi}{T} t+\phi\right)
$$
then we get
$$
\omega=\frac{2 \pi}{T} \text { or } T=\frac{2 \pi}{\omega}
$$
So, clearly, the function represents SHM with a period $T=\frac{2 \pi}{\omega}$. Hence proved.

Topic: Physics, Oscillations
Significance of double fertilization is to

Options:
A gives rise to an endosperm that provides nourishment to the developing embryo.
B increases the viability of the seeds.
C use both the male gametes.
D all of these

The correct answer is:
all of these

Solution:
Double fertilization is characteristic of angiosperms, and it involves syngamy and triple fusion. The fusion of one male gamete with an egg cell is called syngamy, and it results in the formation of diploid zygote. The fusion of second male gamete with a secondary nucleus is called triple fusion, and it results in the formation of triploid primary endosperm nucleus. Double fertilization also increases the viability of the seeds and allows the utilization of both the male gametes produced by the pollen grains.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
$\sin \mathrm{A}+2 \sin 2 \mathrm{~A}+\sin 3 \mathrm{~A}$ is equal to which of the following?
1. $4 \sin 2 \mathrm{~A} \cos ^{2}\left(\frac{\mathrm{A}}{2}\right) \quad[2016-\mathrm{II}]$
$2 \quad 2 \sin 2 \mathrm{~A}\left(\sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}\right)^{2}$
3. $8 \sin \mathrm{A} \cos \mathrm{A} \cos ^{2}\left(\frac{\mathrm{A}}{2}\right)$
Select the correct answer using the code given below:

Options:
A 1 and 2 only
B 2 and 3 only
C 1 and 3 only
D 1,2 and 3

The correct answer is:
1 and 3 only

Solution:
& \text { Let } \mathrm{A}=30^{\circ} \\
\Rightarrow & \sin \mathrm{A}+2 \sin 2 \mathrm{~A}+\sin 3 \mathrm{~A}=\sin 30^{\circ}+2 \sin 60^{\circ}+\sin 90^{\circ} \\
&=\frac{1}{2}+\frac{2 \sqrt{3}}{2}+1=\frac{2 \sqrt{3}+3}{2}
\end{aligned}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2016 (Phase 2)
Six text books numbered $1,2,3,4,5$ and 6 are arranged at random. What is the probability that the text books 2 and 3 will occupy consecutive places ?

Options:
A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{1}{4}$
D $\frac{1}{6}$

The correct answer is:
$\frac{1}{3}$

Solution:
Total number of possible arrangements $\mathrm{n}(\mathrm{s})=6 !$. Since 2 and 3 occupy consecutive places, so, they are grouped together. So, there will be $5 !$ such arrangements. But 2 and 3 can be arranged in themselves in $2 !$ ways
Required Probability $=\frac{5 ! \times 2 !}{6 !}=\frac{2}{6}=\frac{1}{3}$

Topic: Mathematics, Probability, NDA, NDA 2007 (Phase 2)
Some measures of central tendency for $n$ discrete observations are given below:
$\begin{array}{llll}1 . & \text { Arithmetic mean } & 2 . & \text { Geometric mean }\end{array}$
3 . Harmonic mean
4 . Median A desirable property of a measure of central tendency is if every observation is multiplied by $c$, then the measure of central tendency is also multiplied by $c$, where $c>0 .$ Which of the above measures satisfy the property?

Options:
A 1,2 and 3 only
B 1,2 and 4 only
C 3 and 4 only
D $1,2,3$ and 4

The correct answer is:
1,2 and 4 only

Solution:
"If every observation is multiplied by $c$, then the me sure of central tendency is also multiplied by $c$, where $c>0$. Arithmetic mean, Geometric mean and mediar

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 2)
Sometimes a few chromosomes have non-staining secondary constrictions at a constant location. This gives the appearance of a small fragment called the

Options:
A Satellite
B Centromere
C Telomere
D Kinetochore

The correct answer is:
Satellite

Solution:
Sometimes a few chromosomes have non-staining secondary constrictions at a constant location. This gives the appearance of a small fragment called the satellite.

Topic: Biology, Cell - The Unit of Life, NEET
Species with small world populations that are not endangered or vulnerable at present, but are at the risk are called

Options:
A Critically endangered
B Lower risk
C Rare
D Extinct

The correct answer is:
Rare

Solution:
Rare species are organisms that are considered to be very uncommon. They are species with a naturally small population, either localized or thinly scattered. They can include both plants and animals. A rare animal can be an endangered species; most rare organisms are considered threatened. They cannot recover quickly from catastrophic events and face the threat of rapid population decline.

Topic: Biology, Biodiversity and its Conservation, NEET
Spermatogonia develop through:

Options:
A Amitosis
B Mitosis
C Meiosis I
D Meiosis II

The correct answer is:
Mitosis

Solution:
In testis, the immature male germ cells (spermatogonia) produce sperms by spermatogenesis that begins at puberty. The spermatogonia (sing. spermatogonium) present on the inside wall of seminiferous tubules multiply by mitotic division and increase in numbers. Each spermatogonium is diploid and contains 46 chromosomes. Some of the spermatogonia called primary spermatocytes periodically undergo meiosis. A primary spermatocyte completes the first meiotic division (reduction division) leading to formation of two equal, haploid cells called secondary spermatocytes, which have only 23 chromosomes each. The secondary spermatocytes undergo the second meiotic division to produce four equal, haploid
spermatids.

Topic: Biology, Human Reproduction, NEET
Standard cell voltage for the cell $\mathrm{Pb}\left|\mathrm{Pb}^{2+} \| \mathrm{Sn}^{2+}\right| \mathrm{Sn}$ is $-0.01 \mathrm{~V}$. If the cell is to exhibit $E_{\text {cell }}=0$, the value of $\left[\mathrm{Sn}^{2+}\right] /\left[\mathrm{Pb}^{2+}\right]$ should be antilog of $-$

Options:
A $+0.3$
B $0.5$
C $1.5$
D $-0.5$

The correct answer is:
$+0.3$

Solution:
Apply Nernst equation to the reaction $\mathrm{Pb}+\mathrm{Sn}^{2+} \rightarrow \mathrm{Pb}^{2+}+\mathrm{Sn}$
or $\quad E^{\circ}+\frac{0.059}{2} \log \frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}=E_{\text {cell }}$
or $\log \frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}=\frac{0.01 \times 2}{0.059}=0.3\left(\because E_{\text {cell }}=0\right)$
or $\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}=$ antilog $(0.3)$

Topic: Chemistry, Electrochemistry, VITEEE, VITEEE 2018
Standard deviation of $n$ observations $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ is ${\sigma} .$ Then, the standard deviation of the observations $\lambda a_{1}, \lambda a_{2}, \ldots, \lambda a_{n}$ is

Options:
A $\lambda {\sigma}$
B $-\lambda \sigma$
C $|\lambda| \sigma$
D $\lambda^{n} \boldsymbol{\sigma}$

The correct answer is:
$|\lambda| \sigma$

Solution:
Let $x_{i}=a_{1}, a_{2}, a_{3}, \ldots, a_{n}$
and $y_{i}=\lambda a_{1}, \lambda a_{2}, \lambda a_{3}, \ldots, \lambda a_{n-}$
$$
y_{i}=\lambda x_{i}
$$
$\because$
$\begin{aligned} \bar{y} &=\frac{\Sigma y_{i}}{N} \\ &=\frac{\Sigma \lambda x_{i}}{n} \end{aligned}$
$$
\begin{aligned}
&=\lambda \frac{\Sigma x_{i}}{n} \\
&=\lambda \bar{x} \\
\sigma_{y} &=\sqrt{\frac{\Sigma\left(y_{i}-\bar{y}\right)^{2}}{N}} \\
&=\sqrt{\frac{\Sigma(\lambda x_{i}-\lambda \bar{x})^{2}}{n}} \\
&=\sqrt{\frac{\lambda^{2} \Sigma\left(x_{i}-\bar{x}\right)^{2}}{n}} \\
&=\sqrt{\lambda^{2} \sigma_{x}^{2}}=|\lambda| \sigma_{x}=|\lambda| \sigma
\end{aligned}
$$

Topic: Mathematics, Statistics, WBJEE, WBJEE 2016
Standard electrode potential for $\mathrm{Sn}^{4+} / \mathrm{Sn}^{2+}$ couple is $+0.15 \mathrm{~V}$ and that for the $\mathrm{Cr}^{3+} / \mathrm{Cr}$ couple is -0.74 . These two couples in their standard state are connected to make a cell. The cell potential will be

Options:
A $+0.89 \mathrm{~V}$
B $+0.18 \mathrm{~V}$
C $+1.83 \mathrm{~V}$
D $+1.199 \mathrm{~V}$

The correct answer is:
$+0.89 \mathrm{~V}$

Solution:
$$
\begin{aligned}
& E_{\mathrm{Sn}^{4+} / \mathrm{Sn}^{2+}}^{\circ} \mathrm{V} \\
& E_{\mathrm{CI}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V} \\
& E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}(R P)-E_{\text {anode }(\mathrm{RP})}^{\circ} \\
&=0.15-(-0.74) \\
&=+0.89 \mathrm{~V} .
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, NEET, NEET 2011 (Screening)
Standard electrode potential for the cell with cell reaction
$\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$
is $1.1 \mathrm{~V}$. Calculate the standard gibbs energy change for the cell reaction.
(Given $\mathrm{F}=96487 \mathrm{C} \mathrm{mol}^{-1}$ )

Options:
A $-200.27 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $-212.27 \mathrm{~kJ} \mathrm{~mol}-1$
C $-212.27 \mathrm{~J} \mathrm{~mol}^{-1}$
D $-200.27 \mathrm{~J} \mathrm{~mol}^{-1}$

The correct answer is:
$-212.27 \mathrm{~kJ} \mathrm{~mol}-1$

Solution:

Topic: Chemistry, Electrochemistry, NEET, NEET 2022 (Phase 2)
Standard electrode potential of three metals $X$, $Y$ and $Z$ are $-1.2 \mathrm{~V},+0.5 \mathrm{~V}$ and -3.0 V$ respectively. The reducing power of these metals will be

Options:
A $Y \gt X \gt Z$
B $Z \gt X \gt Y$
C $X \gt Y \gt Z$
D $Y \gt Z \gt X$

The correct answer is:
$Z \gt X \gt Y$

Solution:
$\begin{aligned} & E_y^{\circ}=-1.2 \mathrm{~V} \\ & E_x^{\circ}=+0.5 \mathrm{~V} \\ & E_z^{\circ}=-3.0 \mathrm{~V}\end{aligned}$
$\because$ Higher the reduction potential, lesser the reducing power.
$\therefore$ Order of reducing power is $Z \gt X \gt Y$

Topic: Chemistry, Electrochemistry, JIPMER, JIPMER 2014
Standard electrode potential of three metals $X, Y$ and $Z$ are $-1.2 \mathrm{~V},+0.5 \mathrm{~V}$ and $-3.0 \mathrm{~V}$ respectively. The reducing power of these metals will be

Options:
A $Y>X>Z$
B $Z>X>Y$
C $X>Y>Z$
D $Y>Z>X$

The correct answer is:
$Z>X>Y$

Solution:
$$
\begin{aligned}
& E_X^{\circ}=-1.2 \mathrm{~V} \\
& E_Y^{\circ}=+0.5 \mathrm{~V} \\
& E_Z^{\circ}=-3.0 \mathrm{~V} \\
& \therefore Z>X>Y[\because \text { higher the reduction potential, }
\end{aligned}
$$
lesser the reducing power.

Topic: Chemistry, Electrochemistry, NEET, NEET 2011 (Screening)
Standard electrode potential values, E°V for Al³⁺/Al is –1.66 V and that of Tl³⁺/Tl is +1.26 V. Predict about the formation of M³⁺ ion in solution and compare the electropositive character of the two metals.

Options:
A Aluminum (Al) has a higher tendency to form M³⁺ ions in solution.
B Thallium (Tl) has a higher tendency to form M³⁺ ions in solution.
C Both metals have equal tendencies to form M³⁺ ions in solution.
D None of the above statements are correct.

The correct answer is:
Thallium (Tl) has a higher tendency to form M³⁺ ions in solution.

Solution:
Correct Option is : (B)
Thallium (Tl) has a higher tendency to form M³⁺ ions in solution.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Standard enthalpies of formation of O3,CO2,NH3 and HI are 142.2, -393.2, -46.2 and +25.9 kJmol-1 respectively. The order of their increasing stabilities will be

Options:
A O3,CO2,NH3,HI
B CO2,NH3,HI,O3
C O3,HI,NH3,CO2
D NH3,HI,CO2,O3

The correct answer is:
O3,HI,NH3,CO2

Solution:
(i) Exothermic compounds are more stable than endothermic compounds.
(ii) Greater is the amount of heat evolved in the formation of a compound, more will be its stability.

Topic: Chemistry, Thermodynamics (C), NEET
Standard enthalpy and standard entropy changes for the oxidation of ammonia at $298 \mathrm{~K}$ are $-382.64 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and -145.6 $\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. Standard Gibbs energy change for the same reaction at $298 \mathrm{~K}$ is:

Options:
A $-221.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $-339.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $-439.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $-523.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$-339.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=-382.64-298$
$$
\left(\frac{-145.6}{1000}\right)
$$
$$
\begin{aligned}
& =-382.64+43.38 \\
& =-339.3 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
Related Theory
Free energy is an extensive quantity, but the standard free energy is an intensive quantity.

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2004
Standard enthalpy (heat) of formation of liquid water at $25^{\circ} \mathrm{C}$ is around
$$
\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})
$$

Options:
A $-237 \mathrm{~kJ} / \mathrm{mol}$
B $237 \mathrm{~kJ} / \mathrm{mol}$
C $-286 \mathrm{~kJ} / \mathrm{mol}$
D $286 \mathrm{~kJ} / \mathrm{mol}$

The correct answer is:
$-286 \mathrm{~kJ} / \mathrm{mol}$

Solution:
$\left(\Delta H_f^{\circ}\right)_{\text {water }}=-286 \mathrm{~kJ} / \mathrm{mol}$

Topic: Chemistry, Thermodynamics (C), TS EAMCET, TS EAMCET 2017
Standard enthalpy of formation of water is $-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$. When $1800 \mathrm{mg}$ of water is formed from its constituent elements in their standard states the amount of energy liberated is

Options:
A $2 \cdot 86 \mathrm{~kJ}$
B $5 \cdot 72 \mathrm{~kJ}$
C $57 \cdot 2 \mathrm{~kJ}$
D $28 \cdot 6 \mathrm{~kJ}$

The correct answer is:
$28 \cdot 6 \mathrm{~kJ}$

Solution:
(D)
$\mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{H}_{2} \mathrm{O}_{(\ell)} \quad \Delta \mathrm{H}_{\mathrm{f}}^{0}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$
For $18 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$, amount of energy liberated $=286 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\therefore$ For $1.8 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$, amount of energy liberated $=\frac{1.8 \times 286}{18}=28.6 \mathrm{~kJ}$

Topic: Chemistry, Thermodynamics (C), MHT CET, MHT CET 2020 (19 Oct Shift 1)
Standard enthalpy of vaporisation $\Delta_{\mathrm{vap}} H^{\mathrm{s}}$ for water at $100^{\circ} \mathrm{C}$ is $40.66 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The internal energy of vaporisation of water at $100^{\circ} \mathrm{C}$ (in $\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)$ is
(Assume water vapour to behave like an ideal gas).

Options:
A +37.56
B -43.76
C +43.76
D +40.66

The correct answer is:
+37.56

Solution:
$\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \stackrel{100^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{H}_2 \mathrm{O}(g)$
$\Delta_{\text {vap }} H^{\mathrm{s}}=\Delta_{\text {vap }} E^{\mathrm{s}}+\Delta n_g R T$
For the above reaction,
$\begin{gathered}
\Delta n_g=n_p-n_r=1-0=1 \\
\therefore 40.66 \mathrm{~kJ} \mathrm{~mol}^{-1}=\Delta_{\mathrm{vap}} E^{\mathrm{s}}+1 \times 8.314 \\
\quad \times 10^{-3} \times 373 \\
\Delta_{\mathrm{vap}} E^{\mathrm{s}}=40.66 \mathrm{~kJ} \mathrm{~mol}^{-1}-3.1 \mathrm{~kJ} \mathrm{~mol}{ }^{-1} \\
=+37.56 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{gathered}$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2012 (Screening)
Standard entropies of x2,y2 and xy3 are 60, 40 and 50 J K-1mol-1 respectively. For the reaction 12x2+32y2⇌xy3.ΔH=-30kJ to be at equilibrium, the temperature should be

Options:
A 750 K
B 1000 K
C 1250 K
D 500 K

The correct answer is:
750 K

Solution:
ΔS=Sxy3-12Sx2-32Sy2
ΔS=50-12×60-32×40
=-40 J K-1mol-1
At equilibrium ΔG=0
ΔG=ΔH-TΔS
0=-30-(-40×10-3×T)
T=750K

Topic: Chemistry, Thermodynamics (C), NEET
Standard entropies of $\mathrm{X}_{2}, \mathrm{Y}_{2}$ and $\mathrm{XY}_{3}$ are 60 , 30 and $50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ respectively. For the reaction $\frac{1}{2} \mathrm{X}_{2}+\frac{3}{2} \mathrm{Y}_{2} \rightleftharpoons \mathrm{XY}_{3}, \Delta H=-30 \mathrm{~kJ}$ to be at equilibrium, the temperature should be:

Options:
A $750 \mathrm{~K}$
B $1000 \mathrm{~K}$
C $1250 \mathrm{~K}$
D $500 \mathrm{~K}$

The correct answer is:
$750 \mathrm{~K}$

Solution:
$\Delta S$ for the reaction $\frac{1}{2} \mathrm{X}_{2}+\frac{3}{2} \mathrm{Y}_{2} \rightleftharpoons \mathrm{XY}_{3}$
$\Delta S=50-(30+60)=-40 \mathrm{~J}$
For equilibrium $\Delta G=0=\Delta H-T \Delta S$
$T=\frac{\Delta H}{\Delta S}=\frac{-30000}{-40}=750 \mathrm{~K}$

Topic: Chemistry, Thermodynamics (C), VITEEE, VITEEE 2017
Standard entropies of X2,Y2 and XY3 are 60,40 and 50 J/K/mol respectively. At what temperature the following reaction will be at equilibrium

Options:
A 500K
B 750K
C 1000K
D 1250K

The correct answer is:
750K

Solution:
So from this question we know that enthalpy for the reaction is -30 KJ

12X2 + 32Y2 → XY3 ∆H = -30 KJ

So we know from the concept of standard entropies is

∆S = ∆SP - ∆SR∆S = ∆SXY3 -32 ∆SY2-12∆SX2∆S = 50 - 32×40-12×60∆S = -40 JK-1mol-1

So according to the second law of thermodynamics

∆G = ∆H-T∆S

For equilibrium the free Gibb's energy is ∆G = 0

∆H = T∆ST = ∆H∆ST = -30×103J-40JK-1

So the value of temperature is

T = 750 K

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2021 (19 Aug Shift 1)