datavorous/entrance-exam-dataset huggingface dataset:
The boiling point of alkyl halide are higher than those of corresponding alkanes because of

Options:
A dipole-dipole interaction
B dipole-induced dipole interaction
C H-bonding
D None of the above

The correct answer is:
dipole-dipole interaction

Solution:
Due to dipole-dipole interaction the boiling point of alkyl halide is higher as compared to corresponding alkanes.

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2015
The boiling point of water is exceptionally high because

Options:
A there is a covalent bond between $\mathrm{H}$ and $\mathrm{O}$.
B water molecule is linear.
C water molecules associate due to hydrogen bonding.
D water molecule is not linear.

The correct answer is:
water molecules associate due to hydrogen bonding.

Solution:
The high boiling point of water is due to H-bonding.

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2012
The by product of Solvay-ammonia process is

Options:
A $\mathrm{CO}_{2}$
B $\mathrm{NH}_{3}$
C $\mathrm{CaCl}_{2}$
D $\mathrm{CaCO}_{3}$

The correct answer is:
$\mathrm{CaCl}_{2}$

Solution:
$\mathrm{CaCl}_{2}$ is produced as a byproduct in solvay ammonia process.

$$

\begin{array}{l}

\text { (i) } \mathrm{NaCl}+\mathrm{CO}_{2}+\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \\

\longrightarrow \mathrm{NaHCO}_{3}+\mathrm{NH}_{4} \mathrm{Cl}

\end{array}

$$

(I) $\mathrm{CaCO}_{3} \longrightarrow \mathrm{CO}_{2}+\mathrm{CaO}$

$2 \mathrm{NH}_{2} \mathrm{Cl}+\mathrm{CaO} \longrightarrow 2 \mathrm{NH}_{3}+$

$$

\mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{C}

$$

By podust

Topic: Chemistry, s Block Elements, BITSAT, BITSAT 2015
The carbylamine reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
Primary amines (aromatic or aliphatic) on warming with chloroform and alcoholic KOH, gives carbylamine having offensive smell. This reaction is called carbylamine reaction.

Topic: Chemistry, Amines, JEE Main
The carbylamine reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
Primary amines (aromatic or aliphatic) on warming with chloroform and alcoholic KOH, gives carbylamine having offensive smell. This reaction is called carbylamine reaction.

Topic: Chemistry, Amines, BITSAT, BITSAT 2015
The chemical reaction $2 \mathrm{O}_{3} \longrightarrow 3 \mathrm{O}_{2}$ proceeds as follows:
rate law expression should be:

Options:
A $\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}$
B $\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{-1}$
C $\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{2}$
D $\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right]\left[\mathrm{O}_{2}\right]^{2}$

The correct answer is:
$\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{-1}$

Solution:
$\mathrm{O}_{3} \stackrel{\text { Fast }}{\longrightarrow} \mathrm{O}_{2}+\mathrm{O}$
$$
\begin{array}{l}
\mathrm{O}+\mathrm{O}_{3} \stackrel{\text { Slow }}{\longrightarrow} 2 \mathrm{O}_{2} \\
\mathrm{k}=\frac{\left[\mathrm{O}_{2}\right][\mathrm{O}]}{\left[\mathrm{O}_{3}\right]} \\
\text { Rate }=\mathrm{k}^{\prime}\left[\mathrm{O}_{3}\right][\mathrm{O}]
\end{array}
$$
put [O] from (ii)
$$
\mathrm{r}=\frac{\mathrm{k}^{\prime}\left[\mathrm{O}_{3}\right] \mathrm{k}\left[\mathrm{O}_{3}\right]}{\left[\mathrm{O}_{2}\right]}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{-1}
$$
Note: Intermediates are never represented in rate law equation.

Topic: Chemistry, Chemical Kinetics, JEE Main
The chemical reaction $2 \mathrm{O}_{3} \longrightarrow 3 \mathrm{O}_{2}$ proceeds as follows:
rate law expression should be:

Options:
A $\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}$
B $\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{-1}$
C $\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{2}$
D $\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right]\left[\mathrm{O}_{2}\right]^{2}$

The correct answer is:
$\mathrm{r}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{-1}$

Solution:
$\mathrm{O}_{3} \stackrel{\text { Fast }}{\longrightarrow} \mathrm{O}_{2}+\mathrm{O}$
$$
\begin{array}{l}
\mathrm{O}+\mathrm{O}_{3} \stackrel{\text { Slow }}{\longrightarrow} 2 \mathrm{O}_{2} \\
\mathrm{k}=\frac{\left[\mathrm{O}_{2}\right][\mathrm{O}]}{\left[\mathrm{O}_{3}\right]} \\
\text { Rate }=\mathrm{k}^{\prime}\left[\mathrm{O}_{3}\right][\mathrm{O}]
\end{array}
$$
put [O] from (ii)
$$
\mathrm{r}=\frac{\mathrm{k}^{\prime}\left[\mathrm{O}_{3}\right] \mathrm{k}\left[\mathrm{O}_{3}\right]}{\left[\mathrm{O}_{2}\right]}=\mathrm{k}\left[\mathrm{O}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{-1}
$$
Note: Intermediates are never represented in rate law equation.

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2009
The circuit is equivalent to

Options:
A AND gate
B OR gate
C Not gate
D None of these

The correct answer is:
OR gate

Solution:
$\begin{aligned} \mathrm{Y} &=\mathrm{A}+(\overline{\mathrm{A}} \cdot \mathrm{B})=(\mathrm{A}+\overline{\mathrm{A}}) \cdot(\mathrm{A}+\mathrm{B}) \\ &=\mathrm{A}+\mathrm{B} \\ \Rightarrow & \mathrm{OR} \text { gate } \end{aligned}$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2013
The circuit is equivalent to

Options:
A AND gate
B OR gate
C Not gate
D None of these

The correct answer is:
OR gate

Solution:
$\begin{aligned} \mathrm{Y} &=\mathrm{A}+(\overline{\mathrm{A}} \cdot \mathrm{B})=(\mathrm{A}+\overline{\mathrm{A}}) \cdot(\mathrm{A}+\mathrm{B}) \\ &=\mathrm{A}+\mathrm{B} \\ \Rightarrow & \mathrm{OR} \text { gate } \end{aligned}$

Topic: Physics, Semiconductors, JEE Main
The coefficient of friction between the rubber tyres and the road wayis $0.25 .$ The maximum speed with which a car can be driven round a curve of radius $20 \mathrm{~m}$ without skidding is $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)$

Options:
A $5 \mathrm{~m} / \mathrm{s}$
B $7 \mathrm{~m} / \mathrm{s}$
C $10 \mathrm{~m} / \mathrm{s}$
D $14 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$7 \mathrm{~m} / \mathrm{s}$

Solution:
$\mu \mathrm{mg}=\mathrm{mv}^{2} / \mathrm{r}$ or $\mathrm{v}=\sqrt{\mu \mathrm{g} \mathrm{r}}$

or $v=\sqrt{(0.25 \times 9.8 \times 20)}=7 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2020
The coefficient of the middle term in the expansion of $(2+3 \mathrm{x})^{4}$ is:

Options:
A 6
B $5 !$
C $8 !$
D 216

The correct answer is:
216

Solution:
When exponent is $n$ then total number of terms are $\mathrm{n}+1 .$ So, total number of terms in $(2+3 x)^{4}=5$

Middle term is 3rd. $\Rightarrow \mathrm{T}_{3}={ }^{4} \mathrm{C}_{2}(2)^{2} \cdot(3 \mathrm{x})^{2}$

$$

=\frac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2} \times 4 \times 9 \mathrm{x}^{2}=216 \mathrm{x}^{2}

$$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2013
The coefficient of the term independent of $x$ in the expansion of $\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}\right)^{10}$ is

Options:
A $\frac{5}{4}$
B $\frac{7}{4}$
C $\frac{9}{4}$
D None of these

The correct answer is:
$\frac{5}{4}$

Solution:
The $(r+1)$ th term in the expansion of $\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}\right)^{10}$ is given by

$\begin{array}{l}

\mathrm{T}_{r+1}={ }^{10} \mathrm{C}_{r}\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{3}{2 x^{2}}\right)^{r} \\

={ }^{10} \mathrm{C}_{r} \frac{x^{5-\left(\frac{r}{2}\right)}}{3^{5-\left(\frac{r}{2}\right)}} \cdot \frac{3^{r}}{2^{r} x^{2 r}} \\

={ }^{10} \mathrm{C}_{r} \frac{3^{\left(\frac{3 r}{2}\right)-5}-5-\left(\frac{5 r}{2}\right)}{2^{r}} x

\end{array}$

$\mathrm{T}_{r+1}={ }^{10} \mathrm{C}_{r}\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{3}{2 x^{2}}\right)^{r}$ $={ }^{10} \mathrm{C}_{r} \frac{x^{5-\left(\frac{r}{2}\right)}}{3} \cdot \frac{3^{r}}{2^{r} x^{2 r}}$ $={ }^{10} \mathrm{C}_{r} \frac{3^{\left(\frac{3 r}{2}\right)-5}-\left(\frac{5}{2}\right)-\left(\frac{5 r}{2}\right)}{2^{r}} x$ For $\mathrm{T}_{r+1}$ to be independent of $x$, we must have $5-\left(\frac{5 r}{2}\right)=0$ or $r=2$.

Thus, the 3 rd term is independent of $x$ and is equal to ${ }^{10} \mathrm{C}_{2} \frac{3^{3-5}}{2^{2}}=\frac{10 \times 9}{2} \times \frac{3^{-2}}{4}=\frac{5}{4}$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2021
The coefficient of $\mathrm{x}^{4}$ in the expansion of $\left(1+x+x^{2}+x^{3}\right)^{11},$ is

Options:
A 440
B 770
C 990
D 1001

The correct answer is:
990

Solution:
We have coefficient of $x^{4}$ in $\left(1+x+x^{2}+x^{3}\right)^{11}$

$$

\begin{array}{l}

=\text { coefficient of } \mathrm{x}^{4} \text { in }\left(1+\mathrm{x}^{2}\right)^{11}(1+\mathrm{x})^{11} \\

=\text { coefficient of } \mathrm{x}^{4} \text { in }(1+\mathrm{x})^{11}+\text { coefficient } \\

\text { of } \mathrm{x}^{2} \text { in } \\

\text { 11. }(1+\mathrm{x})^{11}+\text { constant term is }{ }^{11} \mathrm{C}_{2} \cdot(1+\mathrm{x})^{11} \\

={ }^{11} \mathrm{C}_{4}+11 .{ }^{11} \mathrm{C}_{2}+{ }^{11} \mathrm{C}_{2}=990

\end{array}

$$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2014
The coefficient of x5 in the expansion of 1+x21+1+x22+…+1+x30 is

Options:
A C551
B C59
C C631-C621
D C530+C520

The correct answer is:
C631-C621

Solution:
1+x21+1+x22+……+1+x30

=1+x211+1+x1+…+1+x9

=1+x211+x10-11+x-1

=1x1+x31-1+x21

∴ Coefficient of x5 in the given expression

= Coefficient of x5 in 1x1+x31-1+x21

= Coefficient of x6 in 1+x31-1+x21

=C631-C621

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2017
The coefficient of $x^{20}$ in the expansion of
$\left(1+x^{2}\right)^{40} \cdot\left(x^{2}+2+\frac{1}{x^{2}}\right)^{-5}$ is

Options:
A ${ }^{30} \mathrm{C}_{10}$
B ${ }^{30} \mathrm{C}_{25}$
C 1
D None of these

The correct answer is:
${ }^{30} \mathrm{C}_{25}$

Solution:
Expression $=\left(1+x^{2}\right)^{40} \cdot\left(x+\frac{1}{x}\right)^{-10}$
$=\left(1+x^{2}\right)^{30} \cdot x^{10}$
The coefficient of $x^{20}$ in $x^{10}\left(1+x^{2}\right)^{30}$ $=$ the coefficient of $\mathrm{x}^{10}$ in $\left(1+\mathrm{x}^{2}\right)^{30}$
$={ }^{30} \mathrm{C}_{5}={ }^{30} \mathrm{C}_{30-5}={ }^{30} \mathrm{C}_{25}$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2012
The coefficient of x-n in 1+xn1+1xn is

Options:
A 0
B 1
C 2n
D 2n

The correct answer is:
1

Solution:
1+xn1+1xn=1+x1+1xn

1+xn1+1xn=1+x+1x+1n

1+xn1+1xn=2+x+1xn

1+xn1+1xn=C0n2+xn·1x+C1n2+xn-1·1x1+…+Cnn·1xn

∴ Coefficient of x-n=Cnn=1

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2018
The coefficient of \(x^{32}\) in the expansion of \(\left(x^4-\frac{1}{x^3}\right)^{15}\) is:

Options:
A \({ }^{-15} \mathrm{C}_3\)
B \({ }^{15} \mathrm{C}_4\)
C \({ }^{-15} \mathrm{C}_5\)
D \({ }^{15} \mathrm{C}_2\)

The correct answer is:
\({ }^{15} \mathrm{C}_4\)

Solution:
We know by Binomial expansion, that \((x+a)^{\mathrm{n}}\)
\(\begin{aligned}
&={ }^{\mathrm{n}} \mathrm{C}_0 x^{\mathrm{n}} \mathrm{a}^0+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}^{\mathrm{n}-1} \cdot \mathrm{a}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^{\mathrm{n}-2} \mathrm{a}^2 \\
&+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^{\mathrm{n}-3} \mathrm{a}^3+{ }^{\mathrm{n}} \mathrm{C}_4 \mathrm{x}^{\mathrm{n}-4} \cdot \mathrm{a}^4+\ldots . .+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^0 a^{\mathrm{n}}
\end{aligned}\)
Given expansion is \(\left(x^4-\frac{1}{x^3}\right)^{15}\)
On comparing we get \(\mathrm{n}=15, \mathrm{x}=\mathrm{x}^4\),
\(a=\left(-\frac{1}{x^3}\right)\)
\(\begin{aligned}
& \therefore\left(x^4-\frac{1}{x^3}\right)^{15}={ }^{15} \mathrm{C}_0\left(\mathrm{x}^4\right)^{15}\left(-\frac{1}{\mathrm{x}^3}\right)^0 \\
& +{ }^{15} \mathrm{C}_1\left(\mathrm{x}^4\right)^{14}\left(-\frac{1}{\mathrm{x}^3}\right)+{ }^{15} \mathrm{C}_2\left(\mathrm{x}^4\right)^{13}\left(-\frac{1}{\mathrm{x}^3}\right)^2 \\
& +{ }^{15} \mathrm{C}_3\left(\mathrm{x}^4\right)^{12}\left(-\frac{1}{\mathrm{x}^3}\right)^3+{ }^{15} \mathrm{C}_4\left(\mathrm{x}^4\right)^{11}\left(-\frac{1}{\mathrm{x}^3}\right)^4+\ldots \\
& \mathrm{T}_{\mathrm{r}+1}={ }^{15} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^4\right)^{15-\mathrm{r}} \cdot\left(-\frac{1}{\mathrm{x}^3}\right)^{\mathrm{r}}=-{ }^{15} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{60-7 \mathrm{r}} \\
& \Rightarrow \mathrm{x}{ }^{60-7 \mathrm{r}}=\mathrm{x}^{32} \Rightarrow 60-7 \mathrm{r}=32 \\
& \Rightarrow 7 \mathrm{r}=28 \Rightarrow \mathrm{r}=4
\end{aligned}\)
So, 5 th term, contains \(\mathrm{x}^{32}\)
\(={ }^{15} C_4\left(x^4\right){ }^{11}\left(-\frac{1}{x^3}\right)^4={ }^{15} C_4 x^{44} x^{-12}\)
\(={ }^{15} \mathrm{C}_4 \mathrm{x}^{32} .\)
Thus, coefficient of \(\mathrm{x}^{32}={ }^{15} \mathrm{C}_4\).

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2010
The color of $\mathrm{CoCl}_{3} .5 \mathrm{NH}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$ is

Options:
A red
B orange
C orange - yellow
D pink

The correct answer is:
pink

Solution:
$\mathrm{CoCl}_{3} .5 \mathrm{NH}_{3} \mathrm{H}_{2} \mathrm{O}$ is pink in colour

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2015
The combination of gates shown below yields

Options:
A OR gate
B NOT gate
C XOR gate
D NAND gate

The correct answer is:
OR gate

Solution:
The final boolean expression is,

$$

X=\overline{(\bar{A} \cdot \bar{B})}=\overline{\bar{A}}+\overline{\bar{B}}=A+B \Rightarrow \text { OR gate }

$$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2014
The combination of gates shown below yields

Options:
A OR gate
B NOT gate
C XOR gate
D NAND gate

The correct answer is:
OR gate

Solution:
The final boolean expression is,

$$

X=\overline{(\bar{A} \cdot \bar{B})}=\overline{\bar{A}}+\overline{\bar{B}}=A+B \Rightarrow \text { OR gate }

$$

Topic: Physics, Semiconductors, JEE Main
The common name of $\mathrm{K}\left[\mathrm{PtCl}_{3}\left(\eta^{2} \cdot \mathrm{C}_{2} \mathrm{H}_{4}\right)\right]$ is

Options:
A potassium salt
B Zeise's salt
C complex salt
D None of these

The correct answer is:
Zeise's salt

Solution:
Zeise's salt is common name of $\mathrm{K}\left[\mathrm{P} \mathrm{C}_{3}\left(\mathrm{\eta}^{2}=\mathrm{C}_{2} \mathrm{H}_{4}\right)\right]$

Topic: Chemistry, Coordination Compounds, BITSAT, BITSAT 2015
The complex number $z=z+i y$ which satisfies

the equation $\left|\frac{z-3 i}{z+3 i}\right|=1,$ lies on

Options:
A the $X$ -axis
B the straight line $y=3$
C a circle passing through origin
D None of the above

The correct answer is:
the $X$ -axis

Solution:
$\left|\frac{z-3 i}{z+3 i}\right|=1 \Rightarrow|z-3 i|=|z+3 i|$

[if $\left|z-z_{1}\right|=\left|z+z_{2}\right|,$ then it is a perpendicular bisector of $z_{1}$ and $z_{2}$ ] Hence, perpendicular bisector of (0,3) and (0,-3) is $X$ -axis.

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2014
The compound which gives the most stable carbonium ion on dehydration is

Options:
A $\mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{OH}$
B $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}$
C $\mathrm{CH}_{2}=\mathrm{CHCH}_{2} \mathrm{CH}_{2} \mathrm{OH}$
D $\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}$

The correct answer is:
$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}$

Solution:

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2014
The compounds $\left[\mathrm{PtCl}_2\left(\mathrm{NH}_3\right)_4\right] \mathrm{Br}_2$ and $\left[\mathrm{PtBr}_2\left(\mathrm{NH}_3\right)_4\right] \mathrm{Cl}_2$ constitutes a pair of

Options:
A
coordination isomers
B
linkage isomers
C
ionization isomers
D
optical isomers

The correct answer is:
ionization isomers

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, Coordination Compounds, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The compressibility of water is $4 \times 10^{-5}$ per unit atmospheric pressure. The decrease in volume of $100 \mathrm{~cm}^{3}$ of water under a pressure of 100 atmosphere will be

Options:
A $0.4 \mathrm{~cm}^{3}$
B $4 \times 10^{-5} \mathrm{~cm}^{3}$
C $0.025 \mathrm{~cm}^{3}$
D $0.004 \mathrm{~cm}^{3}$

The correct answer is:
$0.4 \mathrm{~cm}^{3}$

Solution:
Compressibility $=\frac{1}{\mathrm{~B}}=\frac{\Delta \mathrm{V} / \mathrm{V}}{\mathrm{P}}$.

Here, $\mathrm{P}=100 \mathrm{~atm}$

Compressibility $=4 \times 10^{-5}$ and $\mathrm{V}=100 \mathrm{~cm}^{3}$

Hence, $\Delta \mathrm{V}=0.4 \mathrm{~cm}^{3}$

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2021
The conductivity of a saturated solution of $\mathrm{BaSO}_{4}$ is $3.06 \times 10^{-6} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$ and its equivalent conductance is $1.53 \mathrm{ohm}^{-1} \mathrm{~cm}^{2}$ equiv $^{-1}$. The $\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{BaSO}_{4}$ will be

Options:
A $4 \times 10^{-12}$
B $2.5 \times 10^{-9}$
C $2.5 \times 10^{-13}$
D $4 \times 10^{-6}$

The correct answer is:
$4 \times 10^{-6}$

Solution:
Solublity $=\frac{\text { conductivity } \times 1000}{\Lambda_{\text {eq }}}$
$$
\begin{array}{l}
=\frac{3.06 \times 10^{-6} \times 1000}{1.53}=2 \times 10^{-3} \\
K_{\text {sp }}=\mathrm{S}^{2}=4 \times 10^{-6}
\end{array}
$$

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2009
The conductivity of a weak acid HA of concentration $0.001 \mathrm{~mol} \mathrm{~L}^{-1}$ is $2.0 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$.
If $\Lambda_{\mathrm{m}}^{\circ}(\mathrm{HA})=190 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, the ionization constant $\left(\mathrm{K}_{\mathrm{a}}\right)$ of $\mathrm{HA}$ is equal to _______ x $10^{-6}$.

Options:
A 24
B 48
C 12
D 45

The correct answer is:
12

Solution:
$\Lambda_{\mathrm{m}}=1000 \times \frac{\mathrm{K}}{\mathrm{M}}$
$=1000 \times \frac{2 \times 10^{-5}}{0.001}=20 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
$\Rightarrow \alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}=\frac{20}{190}=\left(\frac{2}{19}\right)$
$\underset{0.001(1-\alpha)}{\mathrm{HA}} \rightleftharpoons \underset{0.001 \alpha}{\mathrm{H}^{+}}+\underset{0.001 \alpha}{\mathrm{A}^{-}}$
$K_{\mathrm{a}}=0.001\left(\frac{\alpha^2}{1-\alpha}\right)=\frac{0.001 \times\left(\frac{2}{19}\right)^2}{1-\left(\frac{2}{19}\right)}$
$=12.3 \times 10^{-6}$

Topic: Chemistry, Electrochemistry, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The cooking time in a pressure cooker is less because:

Options:
A More heat is used
B High pressure cooks the food
C The boiling point of water increases in the cooker
D Heat is uniformly distributed For the reaction $: \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}$

The correct answer is:
The boiling point of water increases in the cooker

Solution:
In pressure cooker, pressure is high thus, the boiling point of water increases, resulting cooking time is less than other open pots.

Topic: Chemistry, States of Matter, BITSAT, BITSAT 2009
The coordinates of the foot of the perpendicular from the point $(2,3)$ on the line $x+y-11=0$ are

Options:
A
(-6,5)
B
(5,6)
C
(-5,6)
D
(6,5)

The correct answer is:
(5,6)

Solution:
Let $(h, k)$ be the coordinates of the foot of the perpendicular from the point $(2,3)$ on the $\mathrm{i}$ $x+y-11=0$. Then, the slope of the perpendicular line is $\frac{k-3}{h-2}$. Again the slope of the given line $x+y-11=0$ is -1
Using the condition of perpendicularity of lines, we have
Since $(h, k)$ lies on the given line, we have,
Solving (i) and (ii), we get $h=5$ and $k=6$. Thus $(5,6)$ are the required coordinates of the foot of the perpendicular.

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The coordinates of the point at which minimum value of Z=7x-8y subject to constraints x+y-20≤0, y≥5, x≥0, y≥0 is attained, is

Options:
A 20,0
B 15,5
C 0,5
D 0,20

The correct answer is:
0,20

Solution:
Feasible region is ABCA and Z=7x-8y

At A0,5, Z=7×0-8×5=-40

At B15,5, Z=715-85=65

At C0,20, Z=70-820=-160

Hence, the minimum value of Z is attained at point 0,20

So, option d is correct.

Topic: Mathematics, Linear Programming, BITSAT, BITSAT 2018
The coordinates of the point where the line through the points $\mathrm{A}(3,4,1)$ and $\mathrm{B}(5,1,6)$ crosses the XY-plane are

Options:
A $\left(\frac{13}{5}, \frac{23}{5}, 0\right)$
B $\left(-\frac{13}{5}, \frac{23}{5}, 0\right)$
C $\left(\frac{13}{5},-\frac{23}{5}, 0\right)$
D $\left(-\frac{13}{5}, \frac{-23}{5}, 0\right)$

The correct answer is:
$\left(\frac{13}{5}, \frac{23}{5}, 0\right)$

Solution:
Equation of the line through the given points is $\frac{x-3}{5-3}=\frac{y-4}{1-4}=\frac{z-1}{6-1}$

$\Rightarrow \frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}$

Any point on this line can be taken as $(3+$ $2 \lambda, 4-3 \lambda, 1+5 \lambda)$

If this point lies on XY-plane then the z-coordinate is zero

$$

\Rightarrow 1+5 \lambda=0 \Rightarrow \lambda=-\frac{1}{5}

$$

Thus the required coordinates of the point are

$$

\left(3-\frac{2}{5}, 4-3\left(-\frac{1}{5}\right), 0\right) \equiv\left(\frac{13}{5}, \frac{23}{5}, 0\right)

$$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2013
The correct bond order in the following species is:

Options:
A $\mathrm{O}_{2}^{2+}<\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{+}$
B $\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{2+}$
C $\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{2+}$
D $\mathrm{O}_{2}^{2+}<\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{-}$

The correct answer is:
$\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{2+}$

Solution:
$\mathrm{O}_{2}^{+}$ ion - Total number of electrons
$(16-1)=15$
E.C.:
$$
\begin{array}{r}
\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \\
\pi 2 p_{x}^{2}=\pi 2 p_{y}^{2} \pi^{*} 2 p_{x}^{1}=\pi^{*} 2 p_{y} \\
\text { Bond order }=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}=\frac{10-5}{2}=\frac{5}{2}=2 \frac{1}{2}
\end{array}
$$
$\mathrm{O}_{2}^{-}$ (Super oxide ion): Total number of
electrons $(16+1)=17$
E.C.:
$$
\begin{array}{l}
\text { \sigma1s }^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \\
\pi 2 p_{x}^{2}=\pi 2 p_{y}^{2} \pi^{*} 2 p_{x}^{2}=\pi^{*} 2 p_{y}^{1} \\
\text { Bond order }=\frac{\left(\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)}{2}=\frac{10-7}{2}=\frac{3}{2}=1 \frac{1}{2}
\end{array}
$$
$\mathrm{O}_{2}^{2+}$ ion: Total number of electrons $=(16-2)=14$
E.C.
$$
\begin{array}{l}
\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2} \sigma 2 p_{z}^{2} \\
\text { Bond order }=\frac{\left(N_{b}-N_{a}\right)}{2}=\frac{10-4}{2}=\frac{6}{2}=3
\end{array}
$$
So bond order: $\mathrm{O}_{2}^{-}<\mathrm{O}_{2}^{+}<\mathrm{O}_{2}^{2+}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2011
The correct nucleophilicity order is

Options:
A $\mathrm{CH}_{3}^{-}<\mathrm{NH}_{2}^{-}<\mathrm{HO}^{-}<\mathrm{F}^{-}$
B $\mathrm{CH}_{3}^{-} \simeq \mathrm{NH}_{2}^{-}>\mathrm{HO}^{-} \simeq \mathrm{F}^{-}$
C $\mathrm{CH}_{3}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{HO}^{-}>\mathrm{F}^{-}$
D $\mathrm{NH}_{2}^{-}>\mathrm{F}^{-}>\mathrm{HO}^{-}>\mathrm{CH}_{3}^{-}$

The correct answer is:
$\mathrm{CH}_{3}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{HO}^{-}>\mathrm{F}^{-}$

Solution:
Nucleophilicity increases with the decrease in electronegativity of the central atom. Since electronegativity follows the order:
$\mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{C} ;$ nucleophilicity of the
concerned group will follow the reverse order i.e.,
$\mathrm{CH}_{3}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{OH}^{-}>\mathrm{F}^{-}$

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2011
The correct order of acidic strength :

Options:
A $\mathrm{Cl}_{2} \mathrm{O}_{7}>\mathrm{SO}_{2}>\mathrm{P}_{4} \mathrm{O}_{10}$
B $\mathrm{K}_{2} \mathrm{O}>\mathrm{CaO}>\mathrm{MgO}$
C $\mathrm{CO}_{2}>\mathrm{N}_{2} \mathrm{O}_{5}>\mathrm{SO}_{3}$
D $\mathrm{Na}_{2} \mathrm{O}>\mathrm{MgO}>\mathrm{Al}_{2} \mathrm{O}_{3}$

The correct answer is:
$\mathrm{Cl}_{2} \mathrm{O}_{7}>\mathrm{SO}_{2}>\mathrm{P}_{4} \mathrm{O}_{10}$

Solution:
Acidic character of oxide $\propto$ Non-metallic nature of element. Non-metallic character increases along the period. Hence order of acidic character is

$\mathrm{Cl}_{2} \mathrm{O}_{7}>\mathrm{SO}_{2}>\mathrm{P}_{4} \mathrm{O}_{10}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2020