datavorous/entrance-exam-dataset huggingface dataset:
Which one of the following is correct order of given isotopes?
I. $T_2>D_2>P_2$ (order of boiling point)
II. $T_2>D_2>P_2$ (order of bond energy)
III. $T_2=D_2=P_2$ (order of bond length)
IV. $T_2 < D_2 < P_2$ (order of reactivity with $\mathrm{Cl}_2$ )

Options:
A
I and II

B
III and IV

C
II, III and IV

D
All of these

The correct answer is:

All of these

Solution:
Protium $(P)$, deuterium $(D)$ and tritium $(T)$ are the three isotopes of hydrogen. These isotopes follow the order in different contexts as shown below :
$T_2>D_2>P_2$ [Order of boiling point (BP)]
$T_2>D_2>P_2$ [Order of bond energy $\left.(\mathrm{BE})\right]$
$T_2=D_2=P_2$ [Order of bond length $\left.(\mathrm{BL})\right]$
$T_2 < D_2 < P_2$ [Order of reactivity with $\mathrm{Cl}_2$ ]
Hence, all the given order are correct.

Topic: Chemistry, Hydrogen, BITSAT, BITSAT 2022
Which one of the following is correct regarding $\Delta \mathrm{H}:$

Options:
A $\quad \Delta \mathrm{H}=\Delta \mathrm{E}+2 \mathrm{RT}$
B $\quad \Delta \mathrm{H}=\Delta \mathrm{E}-2 \mathrm{RT}$
C $\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{RT}$
D $\quad \Delta \mathrm{H}=\Delta \mathrm{E}-\mathrm{RT}$

The correct answer is:
$\quad \Delta \mathrm{H}=\Delta \mathrm{E}-2 \mathrm{RT}$

Solution:
$\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}$
According to thermodynamics's Ist law
$$
\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{nRT}
$$
Where $\Delta \mathrm{H}=$ enthalpy of reaction at constant pressure $\Delta \mathrm{E}=$ heat of reaction at constant volume $\mathrm{R}=$ molar gas constant
$$
\begin{array}{l}
\mathrm{T}=\text { temperature of the reaction } \\
\mathrm{n}=(\text { no. of moles of product })-(\text { no. of moles }
\end{array}
$$
of reactant.) From reaction, $\mathrm{n}=\mathrm{n}_{\mathrm{p}}-\mathrm{n}_{\mathrm{R}}=2-4=-2$
Hence, $\Delta \mathrm{H}=\Delta \mathrm{E}-2 \mathrm{RT}$.

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2009
Which one of the following is the unit vector perpendicular to both \(\vec{a}=-\hat{i}+\hat{j}+\hat{k}\) and \(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\) ?

Options:
A \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\)
B \(\hat{\mathrm{k}}\)
C \(\frac{\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{2}}\)
D \(\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}}\)

The correct answer is:
\(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\)

Solution:
According to question \(\mathrm{a}=-\hat{i}+\hat{j}+\hat{k}\) and \(\mathrm{b}=\hat{i}-\hat{j}+\hat{k}\)
Then, \(\mathrm{a} \times \mathrm{b}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 1 & -1 & 1\end{array}\right|\)
\(\begin{gathered}
=\hat{i}[1+1]-\hat{j}[-1-1]+\hat{k}[1-1]=2(\hat{i}+\hat{j}) \\
\quad \text { and }|\mathrm{a} \times \mathrm{b}|=\sqrt{4+4}=2 \sqrt{2}
\end{gathered}\)
\(\therefore \text { Required unit vector }= \pm \frac{2(\hat{i}+\hat{j})}{2 \sqrt{2}}= \pm \frac{\hat{i}+\hat{j}}{\sqrt{2}}\)

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2010
Which one of the following molecular hydrides acts as a Lewis acid?

Options:
A $\mathrm{NH}_3$
B $\mathrm{H}_2 \mathrm{O}$
C $\mathrm{B}_2 \mathrm{H}_6$
D $\mathrm{CH}_4$

The correct answer is:
$\mathrm{B}_2 \mathrm{H}_6$

Solution:
Boron in $\mathrm{B}_2 \mathrm{H}_6$ is electron deficient

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Which one of the following molecules will form a linear polymeric structure due to hydrogen bonding?

Options:
A $\mathrm{NH}_{3}$
B $\mathrm{H}_{2} \mathrm{O}$
C $\mathrm{HCl}$
D $\mathrm{HF}$

The correct answer is:
$\mathrm{HF}$

Solution:
HF form linear polymeric structure due to hydrogen bonding.

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2021
Which one of the following monomers gives the polymer neoprene on polymerization?

Options:
A $\quad \mathrm{CF}_{2}=\mathrm{CF}_{2}$
B $\mathrm{CH}_{2}=\mathrm{CHCl}$
C $\mathrm{CCl}_{2}=\mathrm{CCl}_{2}$
D \(\mathrm{CH}_2=\stackrel{\huge\mathrm{Cl} \atop \huge |}{\mathrm{C}}-\mathrm{CH}=\mathrm{CH}_2\)

The correct answer is:
\(\mathrm{CH}_2=\stackrel{\huge\mathrm{Cl} \atop \huge |}{\mathrm{C}}-\mathrm{CH}=\mathrm{CH}_2\)

Solution:

Topic: Chemistry, Polymers, BITSAT, BITSAT 2014
Which one of the following pairs is isostructural (i.e., having the same shape and hybridization)?

Options:
A $\left[\mathrm{BCl}_{3}\right.$ and $\left.\mathrm{BrCl}_{3}^{-}\right]$
B $\left[\mathrm{NH}_{3}\right.$ and $\left.\mathrm{NO}_{3}^{-}\right]$
C $\left[\mathrm{NF}_{3}\right.$ and $\left.\mathrm{BF}_{3}\right]$
D $\left[\mathrm{BF}_{4}^{-}\right.$ and $\left.\mathrm{NH}_{4}^{+}\right]$

The correct answer is:
$\left[\mathrm{BF}_{4}^{-}\right.$ and $\left.\mathrm{NH}_{4}^{+}\right]$

Solution:
$\mathrm{BF}_{4}^{-}$ hybridisation $\mathrm{sp}^{3}$, tetrahedral structure.

$\mathrm{NH}_{4}^{+}$ hybridisation $s p^{3},$ tetrahedral structure.

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2014
Which one of the following pairs is not correctly matched?

Options:
A $>\mathrm{C}=\mathrm{O} \rightarrow>\mathrm{CH}_{2}$ (Clemmensen reduction)
B $>\mathrm{C}=\mathrm{O} \rightarrow>\mathrm{CHOH}$ (Wolf-Kishner reduction)
C $-\mathrm{COCl} \rightarrow-$ CHO (Rosenmund reduction)
D $-\mathrm{C} \equiv \mathrm{N} \rightarrow-\mathrm{CHO}$ (Stephen reduction)

The correct answer is:
$>\mathrm{C}=\mathrm{O} \rightarrow>\mathrm{CHOH}$ (Wolf-Kishner reduction)

Solution:
Like clemmensen reduction, Wolf-Kishner reduction involves reduction of $>\mathrm{C}=\mathrm{O}$ to $>\mathrm{CH}_{2},$ of course by different reagent.

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2012
Which one of the following reactions is expected to readily give a hydrocarbon product in good yields?

Options:
A $\mathrm{RCOOK} \frac{\text { Electrolytic }}{\text { oxidation }}$
B $\mathrm{RCOO}^{-} \mathrm{Ag}^{+} \stackrel{\mathrm{Br}_{2}}{\longrightarrow}$
C $\mathrm{CH}_{3} \mathrm{CH}_{3} \frac{\mathrm{Cl}_{2}}{\mathrm{hu}}$
D $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl} \stackrel{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}{\longrightarrow}$

The correct answer is:
$\mathrm{RCOOK} \frac{\text { Electrolytic }}{\text { oxidation }}$

Solution:
Electrolysis of a concentrated aqueous solution of either sodium or potassium salts of saturated carboxylic acids yields higher alkane at anode.

$$

2 \mathrm{RCOOK} \underset{\text { Oxidation }}{\text { Electrolytic }}{\longrightarrow} \underset{\text { Anode }}{2 \mathrm{RCOO}^{-}}+\underset{\text { Cathod }}{2 \mathrm{~K}^{+}}

$$

At anode $2 \mathrm{RCOO}^{-} \rightarrow 2 \mathrm{RCOO}+2 \mathrm{e}^{-} \longrightarrow$

$$

\mathrm{R}-\mathrm{R}+2 \mathrm{CO}_{2}

$$

At Cathode $2 \mathrm{~K}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{~K}$

$$

2 \mathrm{~K}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{KOH}+\mathrm{H}_{2} \uparrow

$$

(Kolbe's Method)

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2013
Which one of the following statements in relation to the hydrogen atom is correct?

Options:
A
$3 s, 3 p$ and $3 d$-orbitals all have the same energy.

B
$3 s$ and $3 p$-orbitals are of lower energy than $3 d$-orbital.

C
$3 p$-orbital is lower in energy than $3 d$-orbital.

D
$3 s$-orbital is lower in energy than $3 p$-orbital.

The correct answer is:

$3 s, 3 p$ and $3 d$-orbitals all have the same energy.

Solution:
The hydrogen atom has the configuration $1 s$ and $3 s, 3 p$ and $3 d$-orbitals will have same energy with respect to $1 s$-orbital.

Topic: Chemistry, Hydrogen, BITSAT, BITSAT 2022
Which one of the following statements is correct?

Options:
A From a mixed precipitate of $\mathrm{AgCl}$ and $\mathrm{AgI}$, ammonia solution dissolves only $\mathrm{AgCl}$.
B Ferric ions give a deep green precipitate on adding potassium ferrocyanide solution.
C On boiling a solution having $\mathrm{K}^{+}, \mathrm{Ca}^{2+}$ and $\mathrm{HCO}_{3}^{-}$ ions we get a precipitate of $\mathrm{K}_{2} \mathrm{Ca}\left(\mathrm{CO}_{3}\right)_{2}$
D Manganese salts give a violet borax bead test in the reducing flame.

The correct answer is:
From a mixed precipitate of $\mathrm{AgCl}$ and $\mathrm{AgI}$, ammonia solution dissolves only $\mathrm{AgCl}$.

Solution:
Ammonia can dissolve ppt. of AgCl only due to formation of complex as given below:
$$
\mathrm{AgCl}+2 \mathrm{NH}_{3} \rightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}
$$

Topic: Chemistry, d and f Block Elements, BITSAT, BITSAT 2010
Which one of the following statements is false?

Options:
A The correct order of osmotic pressure for 0.01 M aqueous solution of each compound is $\mathrm{BaCl}_{2}>\mathrm{KCl}>\mathrm{CH}_{3} \mathrm{COOH}>$ sucrose.
B The osmotic pressure $(\pi)$ of a solution is given by the equation $\pi=\mathrm{MRT}$, where $\mathrm{M}$ is the molarity of the solution.
C Raoult's law states that the vapour pressure of a component over a solution is proportional to its mole fraction.
D Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression.

The correct answer is:
Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression.

Solution:
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m} \times \mathrm{i}$. Since $\mathrm{K}_{\mathrm{f}}$ has different values
for different solvents, hence even if the $\mathrm{m}$ is the same $\Delta \mathrm{T}_{\mathrm{f}}$ will be different.

Topic: Chemistry, Solutions, BITSAT, BITSAT 2009
Which one the following graphs represents the behaviour of an ideal gas

Options:
A
B
C
D

The correct answer is:

Solution:
For an ideal gas $\mathrm{PV}=$ constant i.e., $\mathrm{PV}$ does not vary with $\mathrm{V}$.

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2009
Which one the following removes temporary hardness of water?

Options:
A Slaked lime
B $\quad$ Plaster of Paris
C Epsom
D Hydrolith

The correct answer is:
Slaked lime

Solution:
This method is known as Clark's process. In this method temporary hardness is removed byadding lime water or milk of lime.

$$

\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2} \longrightarrow \underset{\mathrm{ppt} .}{2 \mathrm{CaCO}_{3}} \downarrow+2 \mathrm{H}_{2} \mathrm{O}

$$

Topic: Chemistry, Hydrogen, BITSAT, BITSAT 2015
Which statement is incorrect about peptide bond?

Options:
A $\mathrm{C}-\mathrm{N}$ bond length in proteins is longer than usual $\mathrm{C}-\mathrm{N}$ bond length.
B Spectroscopic analysis shows planar structure of

bond.
C $\mathrm{C}-\mathrm{N}$ bond length in proteins is smaller than usual $\mathrm{C}-\mathrm{N}$ bond length.
D None of these

The correct answer is:
$\mathrm{C}-\mathrm{N}$ bond length in proteins is longer than usual $\mathrm{C}-\mathrm{N}$ bond length.

Solution:
Due to resonance,

bond acquires some double bond character, hence shorter in length.

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2013
Which substance has a dipole moment?

Options:
A CCl4
B CH2Cl2
C C2Cl2
D C2Cl4

The correct answer is:
CH2Cl2

Solution:
CH2Cl2 is similar to CH4 (tetrahedral) but it has three different bond angles.HCH,HCCl,ClCCl, which are close to but not equal to 109.5°. Since, electronegativity of Cl>C>H the bond dipole moments do not cancel and the molecule has a dipole moment.

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2018
Which test among the following is not used for the distinction among 10, 20 and 30 aliphatic amine.

Options:
A Hinsberg's reagent test
B Carbylamine reaction
C Azo dye test
D Action with nitrous acid

The correct answer is:
Azo dye test

Solution:
Hinsberg test for the detection and distinction of primary, secondary, and tertiary amines in a given sample.

Carbylamine reaction is the reaction of a primary amine, chloroform and a base to synthesize isocyanides.

Azo dye test is not used for the detection of primary, secondary and tertiary aliphatic amines.

The Reaction of Amines with Nitrous Acid, primary and secondary amines react with nitrous acid forming arene diazonium salts and N-nitrosoamine respectively.

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2018
Which will not result in the formation of anisole?

Options:
A $\mathrm{PhOH}+\mathrm{CH}_3 \mathrm{Cl} \stackrel{\mathrm{OH}^{-}}{\longrightarrow}$
B $\mathrm{Ph}-\mathrm{Cl}+\mathrm{CH}_3 \mathrm{ONa} \frac{623 \mathrm{~K}}{300 \mathrm{~atm}}$
C

D

The correct answer is:

Solution:
$\left[\mathrm{PtCl}_2\left(\mathrm{NH}_3\right)_4\right] \mathrm{Br}_2$ and $\left[\mathrm{PtBr}_2\left(\mathrm{NH}_3\right)_4\right] \mathrm{Cl}_2$ are ionisation isomers.

Topic: Chemistry, Alcohols Phenols and Ethers, BITSAT, BITSAT 2023 (Memory Based Paper 2)
With 17 consonants and 5 vowels the number of words of four letters that can be formed having two different vowels in the middle and one consonant, repeated or different at each end is

Options:
A 5780
B 2890
C 5440
D 2720

The correct answer is:
5780

Solution:
The two letters, the first and the last of the four lettered word can be chosen in \((17)^2\) ways, as repetition is allowed for consonants. The two vowels in the middle are distinct so that the number of ways of filling up the two places is \({ }^5 \mathrm{P}_2=20\).
The no. of different words \(=(17)^2 \cdot 20=5780\).

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2010
With respect to a rectangular cartesian coordinate system, three vectors are expressed as :
\(\vec{a}=4 \hat{i}-\hat{j}, \vec{b}=-3 \hat{i}+2 \hat{j}\) and \(\vec{c}=-\hat{k} \quad\) where \(\hat{\mathrm{i}}, \hat{\mathrm{j}}, \hat{\mathrm{k}}\) are unit vectors, along the \(X\), \(Y\) and \(Z\)-axis respectively. The unit vector \(\hat{r}\) along the direction of sum of these vector is -

Options:
A \(\hat{\mathrm{r}}=\frac{1}{\sqrt{3}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})\)
B \(\hat{\mathrm{r}}=\frac{1}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})\)
C \(\hat{\mathrm{i}}=\frac{1}{3}(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
D \(\hat{\mathrm{r}}=\frac{1}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)

The correct answer is:
\(\hat{\mathrm{r}}=\frac{1}{\sqrt{3}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})\)

Solution:
\(\begin{aligned}
& \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-\hat{\mathrm{j}}-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}} \\
& \hat{\mathrm{r}}=\frac{\overrightarrow{\mathrm{r}}}{|\mathrm{r}|}=\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{1^2+1^2+(-1)^2}}=\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{3}}
\end{aligned}\)

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2010
Work done in increasing the size of a soap bubble from radius of (3 to 5) cm is nearly (surface tension of soap solution =0.03 Nm-1)

Options:
A 0.2π mJ
B 2π mJ
C 0.4π mJ
D 4π mJ

The correct answer is:
4π mJ

Solution:
Work done = Change in surface energy

r2=5 cm

r1=3 cm

⇒ W=2T×4πr22-r12

=2×0.03×4π52-32×10-4

=0.4π mJ

Topic: Physics, Mechanical Properties of Fluids, BITSAT, BITSAT 2017
Work out the heat change (cal) when $40 \mathrm{~g}$ of He gas at $27^{\circ} \mathrm{C}$ undergoes isothermal and reversible compression from initial pressure of $1 \mathrm{~atm}$ to $10 \mathrm{~atm}\left(\mathrm{R}=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$.

Options:
A $13.818 \mathrm{kcal}$
B $-13.818 \mathrm{kcal}$
C $55.272 \mathrm{kcal}$
D $-55.272 \mathrm{k} \mathrm{cal}$

The correct answer is:
$-13.818 \mathrm{kcal}$

Solution:
$\begin{aligned} \mathrm{q} &=-\mathrm{w}_{\mathrm{rev}}=-\left(-2.303 \mathrm{nRT} \log \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right) \\ &=2.303 \times \frac{40}{4} \times 2 \times 300 \log \frac{1}{10}=-13.82 \mathrm{kcal} \end{aligned}$

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2021