datavorous/entrance-exam-dataset huggingface dataset:
$20 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{NaOH}$ is added to $50 \mathrm{~mL}$ of $0.2 \mathrm{M}$ acetic acid. The $\mathrm{pH}$ of this solution after mixing is $\left(\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5}\right)$

Options:
A $4.5$
B $2.3$
C $3.8$
D 4

The correct answer is:
$4.5$

Solution:
$\mathrm{NaOH}+\mathrm{CH}_{3} \mathrm{COOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}$
$\begin{array}{lllll}\mathrm{m} \mathrm{mol} & 20 & \times 0.2 & 50 \times 0.2 & 0 & 0 \\ \text { added } & =4 & =10 & & \\ & & & \\ \text {-eáction } & 0 & 6 & 4 & 4\end{array}$
$\begin{array}{lcc}\mathrm{m} \text { mol after } & & \\ \text { reaction } 0 & 6 & 4 \\ {\left[\mathrm{CH}_{3} \mathrm{COOH}=\frac{6}{70}\right.} & : & {\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}\end{array}=\frac{4}{70}$
Now since for a basic buffer
$$
\begin{array}{l}
\mathrm{pH}=-\operatorname{logk}_{\mathrm{a}}+\log \frac{[\mathrm{salt}]}{[\mathrm{base}]} \\
\begin{aligned}
\therefore \quad \mathrm{pH} &=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{4 / 70}{6 / 70} \\
&=4.56
\end{aligned}
\end{array}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main
$20 \mathrm{~mL}$ of acetic acid reacts with $20 \mathrm{~mL}$ of ethyl alcohol to form ethyl acetate. The density of acid and alcohol are $1 \mathrm{~g} / \mathrm{mL}$ and $0.7 \mathrm{~g} / \mathrm{mL}$ respectively. The limiting reagent in this reaction is

Options:
A acetic acid
B ethyl alcohol
C acetic acid and ethyl alcohol
D ester

The correct answer is:
ethyl alcohol

Solution:
Mass of $20 \mathrm{~mL}$ of acetic acid $=20 \times 1=20 \mathrm{~g}$
Mass of $20 \mathrm{~mL}$ of ethyl alcohol $=20 \times 0.7=14 \mathrm{~g}$
For the reaction,
$$
\underset{60 \mathrm{~g}}{\mathrm{CH}_{3} \mathrm{COOH}}+\underset{46 \mathrm{~g}}{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}} \underset{-\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}
$$
$60 \mathrm{~g}$ of acetic acid requires $=46 \mathrm{~g}$ of ethyl alcohol $20 \mathrm{~g}$ of acetic acid requires $=\frac{46}{60} \times 20=15.3 \mathrm{~g}$ of ethyl alcohol
Therefore, ethyl alcohol is the limiting reagent.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$20 \mathrm{~mL}$ of methane is completely burnt using $50 \mathrm{~mL}$ of oxygen. The volume of the gas left after cooling to room temperature is

Options:
A $40 \mathrm{~mL}$
B $60 \mathrm{~mL}$
C $30 \mathrm{~mL}$
D $80 \mathrm{~mL}$

The correct answer is:
$30 \mathrm{~mL}$

Solution:
$\therefore$ Volume of gas left after cooling
$=10 \mathrm{mLO}_{2}+20 \mathrm{~mL} \mathrm{CO}_{2}=30 \mathrm{~mL}$
$\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ will be condense to $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ after cooling.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
2.0 molal aqueous solution of an electrolyte X2Y3 is 75% ionised. The boiling point of the solution a 1 atm is Kb(H2O)=0.52 K kg mol-1

Options:
A 274.76Â K
B 377Â K
C 376.4Â K
D 377.16 K

The correct answer is:
377.16 K

Solution:
X2Y3(aq)â2X3+(aq)+3Y2-aq
n=5
ÎTb=i Kb.m=(1+4Î±) Kbm
=(1+4Ã0.75)Ã0.52Ã2
=4.16; Tb=373+4.16=374.76

Topic: Chemistry, Solutions, JEE Main
$20 \%$ of a first order reaction was found to be completed at $10: 00 \mathrm{a} . \mathrm{m}$ at $11.30 \mathrm{a} . \mathrm{m}$. on the same day, $20 \%$ of the reaction was found to be remaining. The half life period in minutes of the reaction is

Options:
A 90
B 45
C 60
D 30

The correct answer is:
45

Solution:
For $t=90 \mathrm{~min}$ (from 10 a.m. to $11: 30$ a.m.)
As $20 \%$ was already completed $\therefore a=80$
$$
\begin{aligned}
& (a-x)=(100-80)=20 \\
& \because \quad k=\frac{2.303}{t} \log \frac{a}{a-x} \\
& k=\frac{2.303}{90} \log \frac{80}{20} \\
& k=\frac{2.303}{90} \log 4=\frac{2.303 \times 0.6020}{90} \\
& k=0.015 \mathrm{~min}
\end{aligned}
$$
$\begin{aligned} & \because t_{1 / 2}=\frac{0.693}{k} \\ & \Rightarrow t_{1 / 2}=\frac{0.693}{0.015}=46.2 \approx 45 \mathrm{~min}\end{aligned}$

Topic: Chemistry, Chemical Kinetics, AP EAMCET, AP EAMCET 2017 (26 Apr Shift 1)
20% of acetic acid is dissociated when its 5 g is added to 500 mL of water. The depression in freezing point of such water is _____×10-3C∘. Atomic mass of C,H and O are 12, 1 and 16 a.m.u. respectively.

[Given : Molal depression constant and density of water are 1.86 K kg mol-1 and 1 g cm-3 respectively.

Options:

The correct answer is:
372

Solution:
Van't Hoff factor (i) for dissociation of acetic acid can be written as follows,

i=1+(n-1)α

n=Number of ions

α=Degree of dissociation

(i=1+0.2(2-1)=1.2

ΔTf=iKfmΔTf=Depression in freezing pointKf=Molar depression constantm=molality

ΔTf=1.2×1.86×5×100060×500

ΔTf=0.372

ΔTf=372×10-3

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (01 Feb Shift 2)
20% of N2O4 molecules are dissociated in a sample of gas at 27° C and 760 torr. Mixture has the density at equilibrium equal to

Options:
A 1.48 g/L
B 1.84 g/L
C 2.25 g/L
D 3.12 g/L

The correct answer is:
3.12 g/L

Solution:
The reaction is
N2O4(g)â2NO2(g)
n = 2 and Î±=20100=0.2
D = 4.6, initial vapour density
d = vapour density at equilibrium
Î±=D-d(n-1)d
0.2=46-d(2-1)d
d = 38.3
Molar mass equilibrium =2Ã38.3=76.6
PM = dRT
Here, d = density of gas mixture
d(mix) =  PM RT  = 1 × 76.6 0.0821 × 300
= 3.12 g/L

Topic: Chemistry, Chemical Equilibrium, JEE Main
20 persons are invited for a party. In how many different ways can they and the host be seated at a circular table, if the two particular persons are to be seated on either side of the host

Options:
A $20 !$
B $2.18 !$
C $18 !$
D None of these

The correct answer is:
$2.18 !$

Solution:
There are $20+1=21$ persons in all. The two particular persons and the host be taken as one unit so that these remain $21-3+1=$ 19 persons to be arranged in 18 ! ways. But the two person on either side of the host can themselves be arranged in 2! ways. Hence there are $2!18!$ ways or $2.18!$ ways.

Topic: Mathematics, Permutation Combination, JEE Main
$22 \mathrm{~g}$ of carbon dioxide at $27^{\circ} \mathrm{C}$ is mixed in closed container with $16 \mathrm{~g}$ of oxygen at $37^{\circ} \mathrm{C}$. If both gases are considered as ideal gases, then the temperature of the mixture is nearly

Options:
A $22.2^{\circ} \mathrm{C}$
B $33.5^{\circ} \mathrm{C}$
C $31.5^{\circ} \mathrm{C}$
D $28.5^{\circ} \mathrm{C}$

The correct answer is:
$31.5^{\circ} \mathrm{C}$

Solution:
Let $T^{\circ} \mathrm{C}$ be the temperature of the mixture.
heat given by $\mathrm{O}_2=$ heat absorbed by $\mathrm{CO}_2$
$\begin{aligned} & \Rightarrow \mu_1 C_{V_1} \Delta T=\mu_2 C_{V_2} \Delta T \\ & \Rightarrow \frac{22}{44}(3 R)(T-27)=\frac{16}{32}\left(\frac{5}{2} R\right)(37-T) \\ & \Rightarrow 3(T-27)=\frac{5}{2}(37-T) \\ & \Rightarrow 6 T-162=185-5 T \\ & \Rightarrow 11 T=347 \\ & \Rightarrow T=31.5 \circ \mathrm{C}\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main
\( 2,3 \)-Dimethyl-2-butene can be prepared by heating which of the following compounds with a strong acid?

Options:
A
B \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} \)
C \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{3} \)
D \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2} \)

The correct answer is:
\( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} \)

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
$240^{\circ}$ is equal to

Options:
A $\left(\frac{4 \pi}{3}\right)^{\circ}$
B $\left(\frac{3 \pi}{4}\right)^{\circ}$
C $\left(\frac{4 \pi}{3}\right)^t$
D $\left(\frac{3 \pi}{4}\right)^t$

The correct answer is:
$\left(\frac{4 \pi}{3}\right)^{\circ}$

Solution:
We know that
$180^{\circ}=\pi^{\circ}$
$240^{\prime}=\left(\frac{\pi}{180} \times 240\right)^{\circ}=\left(\frac{4 \pi}{3}\right)^{\circ}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
\( 2,4 \) and \( 6 \mathrm{~S} \) are the conductances of three conductors. When they are joined in parallel, their equivalent conductance will be

Options:
A \( 12 \mathrm{~S} \)
B \( \frac{1}{12} \mathrm{~S} \)
C \( \frac{12}{11} \mathrm{~S} \)
D \( \frac{11}{12} \mathrm{~S} \)

The correct answer is:
\( 12 \mathrm{~S} \)

Solution:
We know that

R=1S

R1=12 Ω, R2=14 Ω, R3=16 Ω
In parallel combination,

the equivalent resistance is given by

1Rp=1R1+1R2+1R3 =2+4+6=12 Ω

Hence, the equivalent conductance isG=1Rp=12 S

Topic: Physics, Current Electricity, JEE Main
25.0 mL of 0.050 M BaNO32 is mixed with 25.0 mL of 0.020 M NaF.Ksp of BaF2 is 0.5×10–6 at 298K. The ratio of Ba2+F-2 and Ksp is ______.

Options:

The correct answer is:
5

Solution:
Given:

Volume of Ba(NO3)2 solution = 25.0mL Concentration of Ba(NO3)2 solution = 0.050M Volume of NaF solution = 25.0 mL Concentration of NaF solution = 0.020M

Ba+2=25×0.0550=0.025M

F-=25×0.0250=0.01M

Ba+2 F-=25×10-7

Ksp=0.5×10-6=5×10-7

Ratio=Ba+2F-Ksp=25×10-75×10-7=5

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2023 (13 Apr Shift 1)
25190-19190-8190+2190 is divisible by

Options:
A neither 14 nor 34
B 14 but not by 34
C 34 but not by 14
D both 14 and 34

The correct answer is:
34 but not by 14

Solution:
Given,

25190-19190-8190+2190

Now 25190-19190, 8190, 2190 are divisible by 2, as 25190 will give remainder as 1 and 19190 will give remainder as 1 when divided by 2, so combined 25190-19190 will give remainder 0

Now, 25190-8190 & -19190+2190 are divisible by 17, an-bn is divisible by a-b if n is even,
So given number is divisible by 34 as it is divisible by 2 & 17 both,

Now,25190=7×3+4190, dividing by 7 we get,

7×3+4190=4190=1695=295=4·831

Now when 4·7+131 divided by 7 will give remainder as 4·1=4

And 19190=7×3-2190, when divided by 7 we get,

7×3-2190=-2190=2190=2×863=2×7+163

So, when 2×7+163 divided by 7 will give remainder as 2

Similarly 8190=7+1190 will give 1 remainder

And 2190=27+163 will give 1 remainder,

So combining all we get,25190-19190-8190+2190=4-2-1+1=5
⇒Not divisible by 7, so it will not be divisible by 14

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (08 Apr Shift 2)
25 g of an unknown hydrocarbon upon burnig produces 88 g of CO2 and 9g of H2O. This unknown hydrocarbon contains:

Options:
A 24 g of carbon and 1 g of hydrogen
B 18 g of carbon annd 7 g of hydrogen
C 22 g of carbon and 3 g of hydrogen
D 20 g carbon and 5 g of hydrogen

The correct answer is:
24 g of carbon and 1 g of hydrogen

Solution:
2 mole of CO2 contains 2 mole of carbon, Hence weight of carbon =2×12=24 g .
mass of Hydrogen =25-24=1 gram

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2019 (12 Apr Shift 2)
$2.5 \mathrm{~g}$ of the carbonate of a metal was treated with $100 \mathrm{~mL}$ of $1 \mathrm{~N} \mathrm{H}_2 \mathrm{SO}_4$. After the completion of the reaction, the solution was boiled off to expel $\mathrm{CO}_2$ and was then titrated against $1 \mathrm{~N} \mathrm{NaOH}$ solution. The volume of alkali that would be consumed, if the equivalent weight of the metal is 20

Options:
A $50$
B $25$
C $75$
D 100.

The correct answer is:
$50$

Solution:
Equivalent weight of metal carbonate $=20+30=50$
$2.5 \mathrm{~g}$ of metal carbonate $=\frac{2.5}{50}=0.05 \mathrm{eq}$.
Number of equivalent of $\mathrm{H}_2 \mathrm{SO}_4$ would have reacted $=0.05$
Number of equivalent of $\mathrm{H}_2 \mathrm{SO}_4$ taken
$=\frac{100 \times 1}{1000}=0.1$
Number of equivalent of $\mathrm{H}_2 \mathrm{SO}_4$ remains unreacted
$=0.1-0.05=0.05 \mathrm{eq}$.
$\therefore$ Number of equivalent of alkali consumed $=0.05 \mathrm{eq}$.
milli eq. $=$ Normality $\times$ Volume in $\mathrm{mL}$
$\begin{aligned} & \therefore \quad 1.0 \times V=0.05 \times 1000 \\ & V=\frac{0.05 \times 1000}{1.0}=50 \mathrm{~mL}\end{aligned}$

Topic: Chemistry, Redox Reactions, AIIMS, AIIMS 2009
25 mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4 N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N Na2S2O3 was used to reach the end point. The molarity of the household bleach solution is

Options:
A 0.24 M
B 0.48
C 0.024 M
D 0.96 M

The correct answer is:
0.24 M

Solution:
H2O2Bleach25 ml+2I-0.5M30 ml+2H+(from CH3COOH)4 N(10 ml)→I2 +2H2O ...(i)
I2+2Na2S2O30.25 N(48 ml)→Na2S4O6+2NaI ...(ii)
m. mol of I2=12 (m. Moles of Na2S2O3 )
=12×(0.25×48)=6 m.mol
Using equation (i)
1 m. Mol of I2≡1 m. Mol of H2O2
m.mol of H2O2=6 m.mol
Molarity of H2O2=625=0.24M

Topic: Chemistry, Redox Reactions, JEE Main
\(2.5 \mathrm{~g}\) of a non-volatile, non-electrolyte is dissolved in \(100 \mathrm{~g}\) of water at \(25^{\circ} \mathrm{C}\). The solution showed a boiling point elevation by \(2^{\circ} \mathrm{C}\). Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is \(\qquad\) \(\mathrm{mm}\) of \(\mathrm{Hg}\) (nearest integer)
[Given : Molal boiling point elevation constant of water \(\left(\mathrm{K}_{\mathrm{b}}\right)=0.52 \mathrm{~K} . \mathrm{kg} \mathrm{mol}^{-1}\),
\(1 \mathrm{~atm}\) pressure \(=760 \mathrm{~mm}\) of \(\mathrm{Hg}\), molar mass of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\)]

Options:

The correct answer is:
707

Solution:
$\begin{aligned}
& 2=0.52 \times \mathrm{m} \\
& \mathrm{m}=\frac{2}{0.52}
\end{aligned}$
According to question, solution is much diluted
$\begin{aligned}
& \text { so } \frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}}=\frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solvent }}} \\
& \frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}}=\frac{\mathrm{m}}{1000} \times \mathrm{M}_{\text {solvent }} \\
& \Delta \mathrm{P}=\mathrm{P}^{\circ} \times \frac{\mathrm{m}}{1000} \times \mathrm{M}_{\text {solvent }} \\
& \quad=760 \times \frac{\frac{2}{0.52}}{1000} \times 18=52.615 \\
& \mathrm{P}_5=760-52.615=707.385 \mathrm{~mm} \text { of } \mathrm{Hg}
\end{aligned}$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (04 Apr Shift 1)
25 ml of 2 N HCl, 50 ml of 4 N HNO3 and v ml of 2 M H2SO4 are mixed together and the total volume is made up to 1 L after dilution 50 ml of this acid mixture completely reacted with 25 ml of a 1 N Na2CO3 solution. The value of v is

Options:
A 250 ml
B 62.5 ml
C 100 ml
D 125 ml

The correct answer is:
62.5 ml

Solution:
aqeous solution Na2CO3 is basic in nature
Milli equivalents of acids = milli equivalents of Na2CO3
â2Ã25+4Ã50+4ÃxÃ501000=25Ã1
â´x=62.5 ml

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$2.5 \mathrm{ml}$ of $0.4(\mathrm{M})$ weak monoacidic base $\left(\mathrm{k}_{\mathrm{b}}=1 \times 10^{-12}\right.$ at $\left.25^{\circ} \mathrm{C}\right)$ is titrated with $\frac{2}{15}(\mathrm{M}) \mathrm{HCl}$ in water at $25^{\circ} \mathrm{C}$. The concentration of $\mathrm{H}^{+}$at equivalence point is $\left(\mathrm{k}_{\mathrm{w}}=1 \times 10^{-14}\right.$, at $\left.25^{\circ} \mathrm{C}\right)$,

Options:
A $3.7 \times 10^{-13}$ (M)
B $3.2 \times 10^{-7}(\mathrm{M})$
C $3.2 \times 10^{-2}(\mathrm{M})$
D $2.7 \times 10^{-2}(\mathrm{M})$

The correct answer is:
$3.2 \times 10^{-2}(\mathrm{M})$

Solution:
$\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2}$ (At equivalence point)
$0.4 \times 2.5=\frac{2}{15} \times \mathrm{V}_{2}$
$\mathrm{BOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{B} \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(I)$
$\therefore \mathrm{V}_{2}=\frac{0.4 \times 2.5}{\frac{2}{15}}=\frac{0.4 \times 2.5}{2} \times 15=7.5 \mathrm{ml}$
$\therefore$ Concentration of salt $=\frac{0.4 \times 2.5}{2.5+7.5}$
or, Salt of weak base and strong acid = $0.1$
$\therefore\left[\mathrm{H}^{+}\right]=\mathrm{Ch}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}}=\sqrt{\frac{10^{-14} \times 0.1}{10^{-12}}}=\sqrt{10^{-3}}=3.2 \times 10^{-2} \mathrm{M}$
as $C h^{2}=\frac{K_{w}}{K_{b}} \quad \therefore h=\sqrt{\frac{K_{w}}{K_{b} C}}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
$25 \mathrm{~mL}$ of $0.08 \mathrm{~N}$ Mohr salt solution is oxidised by $20 \mathrm{~mL}$ of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ in acid medium. The mass of Mohr's sallt present in 500 cc is

Options:
A $15.68 \mathrm{~g}$
B $19.6 \mathrm{~g}$
C $39.68$
D $39.2 \mathrm{~g}$

The correct answer is:
$15.68 \mathrm{~g}$

Solution:
The reaction of Miohr's salt and $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is as follow:
$6 \mathrm{Fe}^{2+}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14\left[\mathrm{H}^{+}\right] \longrightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}$ $n$-factor of $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ is 6 and $n$-factor of $\mathrm{Fe}^{2+}$ is 1 .
As $n$-factor for $\mathrm{Fe}$ is 1 so,
Molarity = Normality Concentration of Mohr's salt $=0.08 \mathrm{M}$
Moles in $25 \mathrm{~mL}$ of $0.08 \mathrm{M} \mathrm{Mohr's} \mathrm{salt} \mathrm{}$
$$
=0.08 \times \frac{25}{1000}=0.002
$$
$\therefore$ Moles in $500 \mathrm{cc}$ of Mohr's salt
$$
=0.002 \times \frac{500}{25}=0.04
$$
Hence, mass of Mohr's salt in $500 \mathrm{cc}$
$$
=0.04 \times 392=15.68 \mathrm{~g}
$$

Topic: Chemistry, Redox Reactions, JEE Main
$25 \mathrm{ml}$ of a solution of barrium hydroxide on titration with a $0.1$ molar solution of hydrochloric acid gave a litre value of $35 \mathrm{ml}$. The molarity of barium hydroxide solution was

Options:
A
$0.14$

B
$0.28$

C
$0.35$

D
$0.07$

The correct answer is:
$0.07$

Solution:
$\mathrm{Ba}(\mathrm{OH})_2+2 \mathrm{HCI} \rightarrow \mathrm{BaCl}_2+2 \mathrm{H}_2 \mathrm{O}$
Applying Molarity equation, $\frac{\mathrm{M}_1 \mathrm{~V}_1}{\frac{1}{\left(\mathrm{Ba}(\mathrm{OH})_2\right)}}=\frac{\mathrm{M}_2 \mathrm{~V}_2}{2}$ or $25 \times \mathrm{M}_1=\frac{0.1 \times 35}{2} \therefore \mathrm{M}_1=\frac{0.1 \times 35}{2 \times 25}=\frac{0.7}{10}=0.07$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2003
25 mL of silver nitrate solution (1M) is added dropwise to 25 mL of potassium iodide (1.05 M) solution. The ion(s) present in very small quantity in the solution is/are

Options:
A I- only
B K+only
C NO3-only
D Ag+ and I-both

The correct answer is:
Ag+ and I-both

Solution:
To determine the ion(s) present in very small quantity in the solution, we need to consider the chemical reaction that occurs when silver nitrate reacts with potassium iodide. The reaction between AgNO3 and KI is a double replacement reaction and can be represented as follows:

AgNO3 + KI → AgI + KNO3

Millimoles of AgNO3 = 25

Millimoles of KI = 25 × 1.05=26.25

∴KI is in excess & AgI forms negatively charged colloid. (Some Ag+remains in solution) lons Ag+& F-are therefore, present in very small quantity.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (11 Apr Shift 1)
$25 \mathrm{~mL}$ of the given $\mathrm{HCl}$ solution requires $30 \mathrm{~mL}$ of $0.1 \mathrm{M}$ sodium carbonate solution. What is the volume of this $\mathrm{HCl}$ solution required to titrate $30 \mathrm{~mL}$ of $0.2 \mathrm{M}$ aqueous $\mathrm{NaOH}$ solution?

Options:
A $25 \mathrm{~mL}$
B $75 \mathrm{~mL}$
C $50 \mathrm{~mL}$
D $12.5 \mathrm{~mL}$

The correct answer is:
$25 \mathrm{~mL}$

Solution:
$25 \mathrm{~mL}$ of $\mathrm{HCl}$ solution requires $30 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{Na}_{2} \mathrm{CO}_{3}$ solution.

$$

\begin{array}{l}

\because \mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2} \\

\therefore 25 \times \mathrm{N}_{1}=30 \times 0.2\left(0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}=0.2 \mathrm{~N}\right. \\

\left.\mathrm{Na}_{2} \mathrm{CO}_{3}\right) \\

\mathrm{N}_{1}=\frac{6}{25}=0.24 \mathrm{~N}

\end{array}

$$

Now, HCl solution is titrated with $\mathrm{NaOH}$ solution.

$$

\begin{array}{l}

\mathrm{M}_{1} \mathrm{~V}_{1}=\mathrm{M}_{2} \mathrm{~V}_{2} ; 0.24 \mathrm{NHCl}=0.24 \mathrm{M} \mathrm{HCl} \\

\therefore \mathrm{V} \times 0.24 \times 1=30 \times 0.2 \times 1 \Rightarrow \mathrm{V}=25 \mathrm{~mL}

\end{array}

$$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2019 (11 Jan Shift 2)
0.25 mol of formic acid HCO2H is dissolved in enough water to make one litre of solution. The pH of that solution is 2.19. The Ka of formic acid is

Options:
A 6.5×10-3
B 4.3×10-4
C 1.7×10-4
D 5.3×10-2

The correct answer is:
1.7×10-4

Solution:
pH =2.19=-log⁡[H+]
H+=10-2.19= Antilog -2.19
=6.46×10-3M
KaHCOOH=H+HCOO-HCOOH
Here H+=HCOO-=6.46×10-3M
HCOOH=0.25 mol/1L
=0.25 M and KaHCOOH=?
So Ka(HCOOH)=(6.46×10-3)0.252=1.7×10-4 .

Topic: Chemistry, Ionic Equilibrium, JEE Main
25% of the population are smokers. A smoker has 27 times more chances to develop lung cancer then a non-smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is k10.Then the value of k is _____ .

Options:

The correct answer is:
9

Solution:
Given,

In a city, 25% of the population is smoker

So, Psmoker=0.25

Pnon-smoking=0.75

And smoker has 27 times more chance of being diagnosed with lung cancer,

So, if PLung cancernon-smoker=p then PLung cancersmoker=27p,

Now, a person is selected at random and found to be diagnosed with lung cancer,

So, total probability of having a lung cancer will be

PLung cancer=0.25×27p+0.75×p

So the probability of him being smoker (using bayes' theorem) will be PsmokerLung cancer=0.25×27p0.25×27p+0.75×p=2730=910

Now comparing with k10=910,

Then the value of k will be 9.

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (25 Jan Shift 2)
'25 volume' hydrogen peroxide means

Options:
A 1 L marketed solution contains 250 g of H2O2.
B 1 L marketed solution contains 75 g of H2O2.
C 100 mL marketed solution contains 25 g of H2O2.
D 1 L marketed solution contains 25 g of H2O2.

The correct answer is:
1 L marketed solution contains 75 g of H2O2.

Solution:
25 volumes' H2O2 means: 1 L of H2O2 on decomposition give 25 L of O2 (g) at STP.

2H2O2 (l) → 2H2O (l) + O2 (g) 2[2522.4]mole [2522.4]mole

Mass of H2O2 = 2 × 2522.4 × 34 = 75 gram

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2023 (25 Jan Shift 1)
27 identical drops are charged at 22 V each. They combine to form a bigger drop. The potential of the bigger drop will be _____ V.

Options:

The correct answer is:
198

Solution:
Total volume of the drop will remain unchanged,

So, 27×43πr3=43πR3

⇒R=3r

Potential of the small drops, 22=kqr

For bigger drop, potential will be

V=k27q3r=9kqr=9×22=198 V

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (25 Jun Shift 2)
\( 27 \mathrm{~g} \) of \( \mathrm{Al} \) was treated with \( \mathrm{NaOH} \) solution when a white gelatinous precipitate was obtained which upon strong heating gave an oxide. The amount of oxide (in \( g \) ) is

Options:

The correct answer is:
51

Solution:
Following reaction takes place when Al reacts with NaOH.

2Al→NaOH2A(OH)3→∆Al2O3

2 moles of Al (54 g) react with NaOH to produce 1 mole of Al2O3 (102 g).

So,1 (27 g)mole of Al reacts withNaOH to produce 51 g of Al2O3.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
27 similar drops of mercury are maintained at 10 V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ___________ times that of a smaller drop.

Options:

The correct answer is:
243

Solution:
2743πr3=43πR3

R=3r

Potential energy of smaller drop U1=35kq2r

Potential energy of bigger drop : U=35kQ2R

U=35k27q2R

U=35k2727q23r

U=2727335kq2r

U=243U1

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (26 Feb Shift 2)
30.4 kJ of heat is required to melt one mole of sodium chloride and the entropy change at the melting point is 28.4 J K–1 mol–1 at 1 atm. The melting point of sodium chloride is _______ K (Nearest Integer)

Options:

The correct answer is:
1070

Solution:
The equilibrium process can be written as follows,

NaCls⇌NaCllΔS=ΔHTempT=ΔHΔS=30.4×10328.4=1070.4K

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (15 Apr Shift 1)