datavorous/entrance-exam-dataset huggingface dataset:
IUPAC name of following compound is

Options:
A 2–Aminopentanenitrile
B 2–Aminobutanenitrile
C 3–Aminobutanenitrile
D 3–Aminopropanenitrile

The correct answer is:
3–Aminobutanenitrile

Solution:
Identify the longest carbon chain containing the functional group: The longest carbon chain is four carbons long, indicating the prefix "butan-".

Number the carbons in the chain: Begin numbering from the end nearest to the functional group.

Identify and name substituent groups: The amino group (–NH2) is present as a substituent on carbon number 3. It is named as "amino".

Identify the functional group: The functional group present is the nitrile group (–CN).

Combining all parts, the IUPAC name "3-Aminobutanenitrile" is derived.

Hence, the answer is option C.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (30 Jan Shift 2)
IUPAC name of following compound (P) is :

Options:
A 1-Ethyl-5, 5-dimethylcyclohexane
B 3-Ethyl-1,1-dimethylcyclohexane
C 1-Ethyl-3, 3-dimethylcyclohexane
D 1,1-Dimethyl-3-ethylcyclohexane

The correct answer is:
3-Ethyl-1,1-dimethylcyclohexane

Solution:
By IUPAC nomenclature,

The parent chain in this compound is the cyclohexane ring, which consists of six carbon atoms.

The substituents are:

Ethyl group attached to third carbon.

Two methyl groups attached to carbon 1 .

The Numbering must be given the carbons in the cyclohexane ring so that the substituents get the lowest numbers,

Hence, the name is 3-ethyl 1, 1 -dimethylcyclohexane

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 1)
IUPAC name of following hydrocarbon \((X)\) is :

Options:
A 2-Ethyl-3,6-dimethylheptane
B 2,5,6-Trimethyloctane
C 3,4,7-Trimethyloctane
D 2-Ethyl-2,6-diethylheptane

The correct answer is:
2,5,6-Trimethyloctane

Solution:
2,5,6-Trimethyloctane

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (08 Apr Shift 2)
IUPAC name of is :

Options:
A 3-Ethyl-4-methylhex-5-en-2-ol
B 4-Ethyl-3-methylhex-1-en-5-ol
C 3-Methyl-4-ethylhex-1-en-5-ol
D 4-Methyl-3-ethylhex-5-en-2-ol

The correct answer is:
3-Ethyl-4-methylhex-5-en-2-ol

Solution:
According to the priority of functional groups, OH is given top priority. Hence, the numbering starts from right side.

Topic: Chemistry, General Organic Chemistry, JEE Main
IUPAC name of the compound is

Options:
A 2-Chlorocarbonyl ethylbenzenecarboxylate
B 2-Carboxyethylbenzoyl chloride
C Ethyl 2-(chlorocarbonyl)benzenecarboxylate
D Ethyl 1-(chlorocarbonyl)benzenecarboxylate

The correct answer is:
Ethyl 2-(chlorocarbonyl)benzenecarboxylate

Solution:
Main functional group is ester
Ethyl 2-(chlorocarbonyl)benzenecarboxylate

Topic: Chemistry, General Organic Chemistry, JEE Main
Jaundice is a disorder of

Options:
A excretory system
B skin and eyes
C digestive system
D circulatory system

The correct answer is:
digestive system

Solution:
Jaundice is a disorder of digestive system caused due to blockage or inflammation of the bile duct or hyperbilirubinemia. It is mainly characterized by yellow colouration of skin, white portion of eye and urine.

Topic: Biology, Digestion and Absorption, NEET, NEET 2010 (Mains)
Jay fell ill after having sandwich which was kept in the refrigerator. But bacteria/viruses do not survive in cold temperatures, so which response must have the micro-organism adapted to in order to survive

Options:
A Migration
B Conformation
C Suspension
D Regulation

The correct answer is:
Suspension

Solution:
There are four major ways through which an organism can cope with stressful conditions: By regulating the constant body temperature, by changing their body temperature with not maintain constant body temperature by getting conformed, migration to the comfortable area and return when the stressful period gets over and suspension of their activities that help to survive in stressful condition (eg. bacteria, fungi and lower plants form thick-walled spores).

Topic: Biology, Organisms and Populations, NEET
Jet aircrafts fly at altitudes above $30,000 \mathrm{ft}$ where the air is very cold at $-40^{\circ} \mathrm{C}$ and the pressure is $0.28 \mathrm{~atm}$. The cabin is maintained at 1 atm pressure by means of a compressor which exchanges air from outside adiabatically. In order to have a comfortable cabin temperature of $25^{\circ} \mathrm{C}$, we will require in addition -

Options:
A a heater to warm the air injected into the cabin
B an air-conditioner to cool the air injected into the cabin
C neither a heater nor an air-conditioner; the compressor is sufficient
D alternatively heating and cooling in the two halves of the compressor cycle

The correct answer is:
an air-conditioner to cool the air injected into the cabin

Solution:
$\mathrm{PV}^{\gamma}=\mathrm{C}$
$\begin{array}{l}
\mathrm{P}^{1-\gamma} \mathrm{T}^{\gamma}=\mathrm{C} \\
(0.28)^{1-\gamma} \times(233)^{\gamma}=1^{1-\gamma} \times \mathrm{T}^{\gamma} \\
\gamma=\frac{7}{5} \\
(0.28)^{1-7 / 5} \times(233)^{7 / 5}=1^{1-7 / 5} \times \mathrm{T}^{7 / 5} \\
\mathrm{~T}^{7 / 5}=233^{7 / 5} \times(0.28)^{-2 / 5} \\
\mathrm{~T}=233(0.28)^{-2 / 7} \\
\mathrm{~T}=\frac{233}{(0.28)^{2 / 7}}
\end{array}$
$\mathrm{T}$ is coming
more than $298 \mathrm{~K}$ or $25^{\circ} \mathrm{C}$
$\therefore \mathrm{T}$ is more than $25^{\circ} \mathrm{C}$
so to cool it an extra ac is required.

Topic: Physics, Kinetic Theory of Gases, KVPY, KVPY 2011 (SB/SX)
∑k=06C351-k is equal to

Options:
A C451-C445
B C351-C345
C C452-C445
D C352-C345

The correct answer is:
C452-C445

Solution:
Given,

∑r=06C351-k

=C351+C350+C349+C348+C347+C346+C345

Now we know that Crn+Cr+1n=Cr+1n+1 or Crn=Cr+1n+1-Cr+1n

Now using the above formula in given expression we get,

∑r=06C351-k=C351+C350+C349+C348+C347+C346+C446⏟-C445

⇒∑r=06C351-k=C351+C350+C349+C348+C347+C447⏟-C445

⇒∑r=06C351-k=C351+C350+C349+C348+C448-C445

⇒∑r=06C351-k=C351+C350+C349+C449-C445

⇒∑r=06C351-k=C351+C350+C450-C445

⇒∑r=06C351-k=C351+C451-C445

⇒∑r=06C351-k=C452-C445.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (25 Jan Shift 2)
$\sum_{k=0}^{10}{ }^{20} C_k=$

Options:
A $2^{19}+\frac{1}{2}{ }^{20} C_{10}$
B $2^{19}$
C ${ }^{20} C_{10}$
D None of these

The correct answer is:
$2^{19}+\frac{1}{2}{ }^{20} C_{10}$

Solution:
$\sum_{K=0}^{10}{ }^{20} C_k \quad \text { i.e., }{ }^{20} C_0+{ }^{20} C_1+\ldots \ldots .+{ }^{20} C_{10}$
We know that, $(1+x)^n={ }^n C_0+{ }^n C_1 x^1+{ }^n C_2 x^2+\ldots .+{ }^n C_n \cdot x^n$
Put $x=1 ; 2^n={ }^n C_0+{ }^n C_1+{ }^n C_2+\ldots . .+{ }^n C_n$
Put $n=20 ; 2^{20}={ }^{20} C_0+{ }^{20} C_1+{ }^{20} C_2+\ldots \ldots .+{ }^{20} C_{20}$
$\begin{aligned} & 2^{20}+{ }^{20} C_{10}=2\left[{ }^{20} C_0+{ }^{20} C_1+\ldots \ldots+{ }^{20} C_{10}\right] \\
& {\left[{ }^{20} C_0+{ }^{20} C_1+\ldots . .+{ }^{20} C_{10}\right]=2^{19}+\frac{1}{2}{ }^{20} C_{10}} \\
& \sum_{k=0}^{10}{ }^{20} C_k=2^{19}+\frac{1}{2}{ }^{20} C_{10} \end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main
∑k=020Ck202 is equal to

Options:
A C2140
B C2041
C C2040
D C1940

The correct answer is:
C2040

Solution:
∑k=020Ck202=∑k=020Ck20.C20-k20

Sum of suffix is constant so summation will be C2040

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (27 Aug Shift 1)
$\sum_{k=1}^5 \frac{1^3+2^3+\ldots+k^3}{1+3+5+\ldots+(2 k-1)}$ is equal to

Options:
A $22.5$
B $24.5$
C $28.5$
D $32.5$

The correct answer is:
$22.5$

Solution:
$\begin{aligned} & \sum_{k=1}^5 \frac{1^3+2^3+\ldots+k^3}{1+3+5+\ldots .+(2 k-1)} \\ & =\sum_{k=1}^5 \frac{\left(\frac{k(k+1)}{2}\right)^2}{k^2} \\ & \qquad\left[\begin{array}{l}\because n^3=\left(\frac{n(n+1)}{2}\right)^2 \\ \text { and } 1+3+5 \ldots+(2 K-1)=K^2\end{array}\right] \\ & =\sum_{k=1}^5 \frac{(k+1)^2}{4}=\frac{2^2+3^2+4^2+5^2+6^2}{4} \\ & =\frac{4+9+16+25+36}{4}=\frac{90}{4}=22.5\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main
$\sum_{k=1}^6\left[\sin \frac{2 k \pi}{7}-i \cos \frac{2 k \pi}{7}\right]$ is equal to

Options:
A $-1$
B $0$
C $-i$
D $i$

The correct answer is:
$i$

Solution:
We have, $\sum_{k=1}^6\left[\sin \frac{2 k \pi}{7}-i \cos \frac{2 \pi k}{7}\right]$
$\begin{aligned}
& =\sum_{k=1}^6\left[(-i)\left(\cos \frac{2 k \pi}{7}+i \sin \frac{2 \pi k}{7}\right)\right] \\
& =(-i) \sum_{k=1}^6\left(\cos \frac{2 k \pi}{7}+i \sin \frac{2 \pi k}{7}\right)=(-i) \sum_{k=1}^6 \alpha^k
\end{aligned}$
Let $\alpha=\cos \frac{2 \pi k}{7}+i \sin \frac{2 \pi k}{7}$
$=(-i)\left(\alpha+\alpha^2+\alpha^3+\ldots+\alpha^5\right)$
Here, $\alpha+\alpha^2+\alpha^3+\ldots+\alpha^6$ follows G.P. so its sum.
$\mathrm{S}=\frac{\alpha\left(1-\alpha^6\right)}{1-\alpha}=\frac{\alpha-\alpha^7}{1-\alpha}=\frac{\alpha-1}{1-\alpha}=i$
$$
\left[\because \alpha^7=\cos 2 \pi+i \sin 2 \pi=1\right]
$$
Thus, $\alpha=-i(-1)$
$\alpha=i$

Topic: Mathematics, Complex Number, JEE Main
K1 and K2 are the respective equilibrium constant for the reactions
(i) XeF6(g)+H2O(g)⇌XeOF4(g)+2HF(g);K1
(ii) XeO4(g)+XeF6(g)⇌XeOF4(g)+XeO3F2;K2
The equilibrium constant for the reaction
XeO4(g)+2HF(g)⇌XeO3F2(g)+H2O(g) will be

Options:
A K1K22
B K1-K2
C K1K2
D K2K1

The correct answer is:
K2K1

Solution:
(i) XeF6(g)+H2O(g)⇌XeOF4(g)+2HF(g);K1
(ii) XeO4(g)+XeF6(g)⇌XeOF4(g)+XeO3F2;K2
Required equation
XeO4(g)+2HF(g)⇌XeO3F2(g)+H2O(g)
It is obtained by subtracting (i) from (ii), hence equilibrium constant of the required equation will be
K2K1.

Topic: Chemistry, Chemical Equilibrium, JEE Main
$K_1, K_2$ and $K_3$ are the equilibrium constants of the following reactions (I), (II) and (III) respectively:
(I) $\mathrm{N}_2+2 \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}_2$
(II) $2 \mathrm{NO}_2 \rightleftharpoons \mathrm{N}_2+2 \mathrm{O}_2$
(III) $\mathrm{NO}_2 \rightleftharpoons \frac{1}{2} \mathrm{~N}_2+\mathrm{O}_2$
The correct relation from the following is

Options:
A
$K_1=\frac{1}{K_2}=\frac{1}{K_3}$

B
$K_1=\frac{1}{K_2}=\frac{1}{\left(K_3\right)^2}$

C
$K_1=\sqrt{K_2}=K_3$

D
$K_1=\frac{1}{K_2}=K_3$

The correct answer is:
$K_1=\frac{1}{K_2}=\frac{1}{\left(K_3\right)^2}$

Solution:
(I) $\mathrm{N}_2+2 \mathrm{O}_2 \stackrel{K_1}{\rightleftharpoons} 2 \mathrm{NO}_2$
$$
K_1=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]^2}
$$
(II) $2 \mathrm{NO}_2 \underset{K_2}{\rightleftharpoons} \mathrm{N}_2+2 \mathrm{O}_2$
$$
K_2=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]^2}{\left[\mathrm{NO}_2\right]^2}
$$
(III)
$$
\begin{aligned}
& \mathrm{NO}_2 \underset{K_3}{\rightleftharpoons} \frac{1}{2} \mathrm{~N}_2+\mathrm{O}_2 \\
& K_3=\frac{\left[\mathrm{N}_2\right]^{1 / 2}\left[\mathrm{O}_2\right]}{\left[\mathrm{NO}_2\right]} \\
& \therefore\left(K_3\right)^2=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]^2}{\left[\mathrm{NO}_2\right]^2}
\end{aligned}
$$
$\therefore$ from equation (i), (ii) and (iii)
$$
K_1=\frac{1}{K_2}=\frac{1}{\left(K_3\right)^2}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2012 (07 May Online)
K2Cr2O7 paper acidified with dilute H2SO4 turns green when exposed to

Options:
A Carbon dioxide
B Sulphur trioxide
C Hydrogen sulphide
D Sulphur dioxide

The correct answer is:
Sulphur dioxide

Solution:
Carbon dioxide and sulphur trioxide cannot undergo oxidation. Potassium dichromate is reduced to chromic sulphate in the presence of sulphur dioxide. The orange solution of potassium dichromate becomes green colour chromic sulphate.

K2Cr2O7 + 3SO2 + H2SO4 → K2SO4 + Cr2(SO4)3 + H2O

Hence, the gas is sulphur dioxide.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (24 Jan Shift 2)
K2HgI4 is 55% ionized in aqueous solution. The value of Van't Hoff factor is

Options:
A 2.1
B 4.3
C 1.9
D 3.7

The correct answer is:
2.1

Solution:
K2HgI4â2K++HgI42-
⇒i = 1 +  n−1 α
⇒i = 1 +  0.55 3-1  (∵n=3)
∴ i=2.1

Topic: Chemistry, Solutions, JEE Main
$\mathrm{K}_{2} \mathrm{HgI}_{4}$ is $40 \%$ ionised in aqueous solution. The value of its van't Hoff factor (i) is:

Options:
A 1.6
B 1.8
C 20
D 22

The correct answer is:
1.8

Solution:
$\mathrm{K}_{2} \mathrm{Hgl}_{4} \rightleftharpoons 2 \mathrm{~K}^{+}+\left[\mathrm{Hgl}_{4}\right]^{2-} ; \mathrm{n}=3$

$$

\begin{aligned}

\because & \alpha=\frac{i-1}{n-1} ; 0.4=\frac{\mathrm{i}-1}{3-1} \\

& i=1.8

\end{aligned}

$$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Ka for butyric acid C3H7COOH is 2×10-5. ThepH of 0.2M solution of butyric acid is____×10-1. (Nearest integer) [Given log2=0.30]

Options:

The correct answer is:
27

Solution:
Ka of Butyric acid ⇒2×10-5 pKa=-logKa = 4.7

pH of weak acid solution

pH=12pKa-12logC

=124·7-12log0.2

=2.35+0.35=2.7

pH=27×10-1

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2022 (28 Jul Shift 1)
Ka for CH3COOH is 1.8×10-5 and Kb for NH4OH is 1.8×10-5. The pH of ammonium acetate solution will be

Options:

The correct answer is:
7

Solution:
Ka of CH3COOH=1.8×10-5

Ka of NH4OH=1.8×10-5

We know that pH of solution of the salt made up of weak acid CH3COOH and weak base NH4OH can be found using formula

pH=7+12pKa-pKb

pKa=-logKa
=-log1.8×10-5=-log 1.8+log 10-5=-0.2553-5=+4.7447pKb=-logKb=-log1.8×10-5=4.7447pH=7+124.7447-4.7447 =7+0 =7

If Ka=Kb, then pH of salt is always 7.
Therefore, the pH of given solution is 7.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2024 (01 Feb Shift 1)
$K_a$ for $\mathrm{HCN}$ is $5 \times 10^{-10}$ at $25^{\circ} \mathrm{C}$. For maintaining a constant $\mathrm{pH}=9$, the volume of $5 \mathrm{M} \mathrm{KCN}$ solution required to be added to $10 \mathrm{~mL}$ of $2 \mathrm{M} \mathrm{HCN}$ solution is

Options:
A $4 \mathrm{~mL}$
B $2.5 \mathrm{~mL}$
C $2 \mathrm{~mL}$
D $6.4 \mathrm{~mL}$

The correct answer is:
$2 \mathrm{~mL}$

Solution:
$\mathrm{pH}=\mathrm{p} K_a+\log \frac{[\text { Salt }]}{[\text { Acid }]}$
Let the volume of $\mathrm{KCN}$ solution required be $V \mathrm{~mL}$
$\therefore \quad[\mathrm{KCN}]=\frac{5 \times V}{V+10}$ and $[\mathrm{HCN}]=\frac{10 \times 2}{V+10}$
Now from eqn. (i),
$\begin{aligned} & \mathrm{pH}=-\log \left(5 \times 10^{-10}\right)+\log \left[\frac{5 \times V}{V+10} / \frac{10 \times 2}{V+10}\right] \\ & 9=-\log \left(5 \times 10^{-10}\right)+\log \frac{V}{4}\end{aligned}$
On solving, $V=1.99 \approx 2 \mathrm{~mL}$

Topic: Chemistry, Ionic Equilibrium, AIIMS, AIIMS 2017
Ka for HCN is 5×10-10 at 25°C . For maintaining a constant pH of 9, the volume in ml of 5 M KCN solution required to be added to 10 ml of 2 M HCN solution is

Options:

The correct answer is:
2.00

Solution:
Suppose, volume of KCN added = x ml
Total volume of solution after mixing = (10 + x) ml
Molarity of HCN in the final solution
=M1V1V2=2Ã10(10+x)=20(10+x)M
Molarity of HCN in the final solution
=M1V1V2=5Ãx(10+x)=5x(10+x)M
pH =-logâ¡ka+logâ¡[KCN][HCN]
9 = −log  (5×10 −10 ) + log  5x 10 + x 20 10 + x
On solving, x = 2 mL

Topic: Chemistry, Ionic Equilibrium, JEE Main
Ka for HCN is 5×10-10 at 25°C . For maintaining a constant pH of 9, the volume of 5 M KCN solution required to be added to 10 mL of 2 M HCN solution is

Options:
A 4 mL
B 8 mL
C 2 mL
D 9 mL

The correct answer is:
2 mL

Solution:
It is a salt of weak acid and strong base so acidic buffer is formed here, hence
pH=pKa+log10[Salt][Acid]
pH=-log5×10-10+log5×VV+1010×2V+10
or 9=-log5×10-10+logV4
So V = 2 mL

Topic: Chemistry, Ionic Equilibrium, NEET
Ka for HCN is 5 x 10 - 10 at 25o C. For maintaining a constant pH of 9, the volume of 5 M KCN solution required to be added to 10 mL of 2 M HCN solution is:

Options:
A 4 mL
B 7.95 mL
C 9.3 mL
D 2 mL

The correct answer is:
2 mL

Solution:
On mixing the given solutions, acidic buffer will be formed. Thus applying Henderson equation to find the pH-
pH = pKa + log10 [CN-][HCN]
pH = - log 5 Ã 1 0 - 1 0 + log 5 Ã V / V + 1 0 1 0 Ã 2 / V + 1 0
9 = - log 5 Ã 1 0 - 1 0 + log V 4
9 = 1 0 - 0 . 7 + log V 4
- 0 . 3 = log V 4
â´ log V 4 = - 0 . 3 = log 1 2
V 4 = 1 2
â´ V = 2 mL.

Topic: Chemistry, Ionic Equilibrium, JEE Main
$\mathrm{K}_{\mathrm{a}}$ of $\mathrm{H}_2 \mathrm{O}_2$ is of the order of

Options:
A $10^{-12}$
B $10^{-14}$
C $10^{-16}$
D $10^{-10}$

The correct answer is:
$10^{-12}$

Solution:
$K_{\mathrm{a}}$ of $\mathrm{H}_2 \mathrm{O}_2=1.55 \times 10^{-12}$

Topic: Chemistry, Hydrogen, JEE Main
Kb for CH2ClCOO- is 6.4Ã10-12 . The pH of 0.1 M CH2ClCOONa in water is :

Options:
A 7.9
B 6.9
C 1.9
D 12.1

The correct answer is:
7.9

Solution:
CH 2 ClCOO - + H 2 O â CH 2 ClCOOH + OH -
OH â â=âchâ=âCâ K H c â=â K H Â·c â=â K w âÃâc K a
= K b Ã c = 6.4 Ã 1 0 - 1 2 Ã 0.1
= 8 Ã 1 0 - 7
â´ pOH = 6.1
pH = 7.9

Topic: Chemistry, Ionic Equilibrium, JEE Main
$K_{c}$ for the the reaction, $\left[\mathrm{Ag}(\mathrm{CN})_{2}\right] \rightleftharpoons \mathrm{Ag}^{+}+2 \mathrm{CN}^{-},$

the equillibrium constant at $25^{\circ} \mathrm{C}$ is $4.0 \times 10^{-19}$ then the silver ion concentration in a solution which was originally 0.1 molar in $\mathrm{KCN}$ and 0.03 molar in $\mathrm{AgNO}_{3}$ is:

Options:
A $7.5 \times 10^{18}$
B $7.5 \times 10^{-19}$
C $7.5 \times 10^{19}$
D $7.5 \times 10^{-18}$

The correct answer is:
$7.5 \times 10^{-18}$

Solution:
$\left.\begin{array}{c}2 \mathrm{KCN}+\mathrm{AgNO}=\left[\underset{0.1}{0.03}=\left[\begin{array}{c}\mathrm{Ag}(\mathrm{CN})_{2} \\ 0\end{array}\right]^{-}\right. \\ (0.1-0.06) & 0\end{array}\right] \begin{array}{c}\mathrm{KNO}_{3}+\mathrm{K}_{0}^{+} \\ (0.03)\end{array}$

$\therefore\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}=\mathrm{Ag}^{+}+\underset{\mathrm{a}}{0.03} \mathrm{a} \quad 2 \mathrm{CN}$

$(0.03-\mathrm{a}) \quad \mathrm{a} .04(\mathrm{left} \mathrm{from} \mathrm{KCN}$

$\quad 0.04 \approx 0.04$

$\therefore \quad \mathrm{K}_{\mathrm{c}}=4 \times 10^{-19}=\frac{(0.04)^{2} \times \mathrm{a}}{0.03}$

$\mathrm{a}=7.5 \times 10^{-18}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
$\mathrm{K}_{\mathrm{f}}$ for water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. If your automobile radiator holds $1.0 \mathrm{~kg}$ of water, how many grams of ethylene glycol $\left(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2\right)$ must you add to get the freezing point of the solution lowered to $-2.8^{\circ} \mathrm{C}$ ?

Options:
A
$72 \mathrm{~g}$

B
$93 \mathrm{~g}$

C
$39 \mathrm{~g}$

D
$27 \mathrm{~g}$

The correct answer is:
$93 \mathrm{~g}$

Solution:
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \cdot \mathrm{m}$
$2.8=1.86 \times \frac{\mathrm{wt}}{62} \times \frac{1000}{1000}$
$\mathrm{Wt}=93 \mathrm{~g}$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2012 (Offline)
$\mathrm{K}_f$ value of $\mathrm{H}_2 \mathrm{O}$ is 1.86 . The value of $\Delta T_f$ for $0.1 \mathrm{~m}$ solution of non-volatile solute is

Options:
A 18.6
B 0.186
C 1.86
D 0.0186

The correct answer is:
0.186

Solution:
$\Delta T_f=K_f \times m=1.86 \times 0.1=0.186$

Topic: Chemistry, Solutions, JEE Main
$K$ is the force constant of a spring. The work done in increasing its extension from $I_1$ to $I_2$ will be

Options:
A $K\left(I_2-I_1\right)$
B $\frac{K}{2}\left(I_2+I_1\right)$
C $K\left(I_2^2-I_1^2\right)$
D $\frac{K}{2}\left(I_2^2-I_1^2\right)$

The correct answer is:
$\frac{K}{2}\left(I_2^2-I_1^2\right)$

Solution:
At extension $l_1$, the stored energy $=\frac{1}{2} K l_1^2$
At extension $l_2$, the stored energy $=\frac{1}{2} \mathrm{Nl}_2^2$
Work done in increasing its extension from $l_1$ to $l_2=\frac{1}{2} K\left(l_2^2-l_1^2\right)$

Topic: Physics, Mechanical Properties of Solids, JEE Main
KMnO4 reacts with KI, in basic medium to form I2 and MnO2 . When 250 mL of 0.1 M KI solution is mtextxed with 250 mL of 0.02 M KMnO4 in basic medium, what is the number of moles of I2 formed?

Options:
A 0.015
B 0.0075
C 0.005
D 0.01

The correct answer is:
0.0075

Solution:
Number of mtextlli equivalents of MnO4-=0.02×3×250=15
Number of mtextlli equivalents of l2=0.1×1×250=25
Thus, here MnO4- is limtextting reagent.
∴ Number of mtextlli equivalents of l2 formed = Number of mtextlli equivalent of MnO4-=15
Or number of equivalent of l2 formed =151000=0.015
∴ Number of moles of l2 formed =0.0152
=0.0075

Topic: Chemistry, Redox Reactions, JEE Main
Kp for the reaction PCl5g⇌PCl3g+Cl2g at 250 o C is 0.82 . Calculate the degree of dissociation at given temperature under a pressure of 5 atm . What will be the degree of dissociation if the equilibrium pressure is 10 atm at same temperature.

Options:
A 27.5%
B 23%
C 35.5%
D 40%

The correct answer is:
27.5%

Solution:
KP=x1+xP×x1+xP1-x1+xP
KP=x21-x2P=0.82
x2=0.825.82
x=0.375
Now new pressure is 10
KP=y2×P1-y2 , 0.82=y21-y2×10
y=0.275 , y=27.5%

Topic: Chemistry, Chemical Equilibrium, JEE Main
Ksp of Al(OH)3=10-36
and E Al 3+ /Al o â=ââ1.66âV
Reduction potential of Al 3+ /Al couple at pHâ=â12 and 298âK is

Options:
A 1.07 V
B 2.25 V
C −1.07 V
D â2.25âV

The correct answer is:
−2.25 V

Solution:
Al3++3e-→Al
E=Eo-0.0593log⁡1Al3+
at pH = 12⇒pOH = 14−12 = 2⇒ OH −  = 0.01 M
Al+3=10-36[OH-]3=10-36(0.01)3=10-30
∴E=-1.66-0.0593log⁡110-30
= −1.66− 0.059 3 × 30 = −1.66−0.59 = −2.25 V

Topic: Chemistry, Electrochemistry, JEE Main
Kα wavelength emitted by an atom of atomic number Z=11 is λ. Find the atomic number for an atom that emits Kα radiation with wavelength 4λ.

Options:

The correct answer is:
6

Solution:
1 λ ∝ z-1 2
∴ λ 1 λ 2 = z 2 - 1 2 z 1 - 1 2
1 4 = Z 2 - 1 1 1 - 1 2
Solving this, we get, Z 2 = 6

Topic: Physics, Atomic Physics, JEE Main
KClO3 + 6FeSO4 + 3H2SO4 → KCl + 3Fe2(SO4)3 + 3H2O

The above reaction was studied at 300 K by monitoring the concentration of FeSO4 in which initial concentration was 10 M and after half an hour became 8.8 M. The rate of production of Fe2(SO4)3 is _______ × l0–6 mol L–1 s –1

(Nearest integer)

Options:

The correct answer is:
333

Solution:
KClO3 + 6FeSO4 + 3H2SO4 → KCl + 3Fe2(SO4)3 + 3H2O

From the above equation, rate of decomposition FeSO4 and rate of production of Fe2SO43 can be written as follows,

16×Rate of decompositionFeSO4=13×Rate of productionFe2SO43

Rate of decomposition of FeSO4

Change in concentrationtime in seconds=(10-8.8)30×60

=1.21800 mol L-1 s-1

Rate of production of Fe2SO43

=0.61800=3.33×10-4

=333.3×10-6 mol-1 s-1

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Keeping in view the 'fluid mosaic model' for the structure of cell membrane, which one of the following statements is correct with respect to the movement of lipids and proteins from one lipid monolayer to the other (described as flip-flop movement)?

Options:
A Both lipids and proteins can flip-flop
B While lipids can rarely flip-flop, proteins cannot
C While proteins can flip-flop, lipids cannot
D Neither lipids, nor proteins can flip-flop

The correct answer is:
While lipids can rarely flip-flop, proteins cannot

Solution:
Singer and Nicolson (1972) suggested the widely accepted fluid mosaic model of biological membranes. According to this model the plasma membrane contains a bimolecular lipid layer, both surfaces of which are interrupted by protein molecules. Some proteins are attached at the polar surface of the lipid (ie, called extrinsic protein) while, other (called intrinsic proteins) either partially penetrate the bilayer or span the membrane entirely to stick out on both sides (called transmembrane proteins). Mobility of membrane proteins due to fluid property of lipid bilayer was demonstrated by classical experiment of D Frye and M Edidin (1970). The movement of lipid molecules from one lipid monolayer to another monolayer is called flip-flop-movement. Flip-flop movement is rarely found in lipid molecules whereas it remain absent in protein molecules.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2008 (Screening)
Keeping in view the 'fluid mosaic model' for the structure of cell membrane, which one of the following statements is correct w.r.t. the movement of lipids and proteins from one lipid monolayer to the other (described as flip-flop movement)?

Options:
A Neither lipids nor proteins can flip-flop
B Both lipids and proteins can flip-flop
C While lipids can rarely flip-flop, proteins cannot
D While proteins can flip-flop, lipids cannot

The correct answer is:
While lipids can rarely flip-flop, proteins cannot

Solution:
It is exclusive property of lipids shown during changing temperature of cellular environment.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2008 (Mains)