datavorous/entrance-exam-dataset huggingface dataset:
Food preservatives prevent spoilage of food due to microbial growth. The most commonly used preservatives includes

Options:
A Table salt and sugar
B Vegetable oils and sodium benzoate (C6H5COONa) .
C Salts of sorbic acid and propanoic acid.
D All of these

The correct answer is:
All of these

Solution:
The most commonly used preservatives include table salt, sugar, vegetable oils and sodium benzoate, C6H5COONa . Sodium benzoate is used in limited quantities and is metabolised in the body. Salts of sorbic acid and propanoic acid are also used as preservatives.

Topic: Chemistry, Chemistry in Everyday Life, NEET
"Foolish seedling" disease of rice led to the discovery of:

Options:
A ABA
B 2, 4-D
C IAA
D GA.

The correct answer is:
GA.

Solution:
Gibberellins (GA) were first isolated from the fungus Gibberella fujikuroi, the causal organism of foolish seedling disease of rice plants, by Kurasawa in Japan in 1926. Abscisic acid was first identified and isolated chemically from cotton plants. Auxin was discovered during the study of bending of the coleoptile of canary grass (Phalaris canariensis) towards light. 2,4-D is an herbicide that kills plants by changing the way certain cells grow.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2007
Fool’s gold is:

Options:
A FeS2
B ZnCl2
C CuFeS2
D Cu2S

The correct answer is:
FeS2

Solution:
FeS2 is known as fool’s gold.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

Options:
A [Co(H2O)3Cl3].3H2O
B [Co(H2O)6]Cl3
C [Co(H2O)5Cl]Cl2.H2O
D [Co(H2O)4Cl2]Cl.2H2O

The correct answer is:
[Co(H2O)3Cl3].3H2O

Solution:
The complex giving least number of ions will have lowest depression in freezing point and therefore highest freezing point. Hence, option 1 is correct. (Van’t Hoff factor = 1)
Freezing point order is options (A > D > C > B).

Topic: Chemistry, Solutions, NEET
For 1 mole of gas, the average kinetic energy is given as E. The urms of gas is

Options:
A 2EM1/2
B 3EM1/2
C 2E2M1/2
D 3E2M1/2

The correct answer is:
2EM1/2

Solution:
Average kinetic energy
E=32RT
urms=3RTM
=2EM .

Topic: Chemistry, States of Matter, NEET
For a black body at temperature $727^{\circ} \mathrm{C}$, its radiating power is $60 \mathrm{~W}$ and the temperature of surrounding is $227^{\circ} \mathrm{C}$. If temperature of the black body is changed to $1227^{\circ} \mathrm{C}$, then its radiating power will be:

Options:
A $304 \mathrm{~W}$
B $320 \mathrm{~W}$
C $240 \mathrm{~W}$
D $120 \mathrm{~W}$

The correct answer is:
$320 \mathrm{~W}$

Solution:
We can use the formula
$\begin{aligned}
& P \propto\left(T_4-T_0^4\right) \\
& \therefore \frac{P_2}{P_1}=\frac{(1500)^4-(500)^4}{(1000)^4-(500)^4} \\
& =\frac{500^4\left(3^4-1\right)}{500^4\left(2^4-1\right)} \\
& \therefore \frac{P_2}{60}=\frac{80}{15} \Rightarrow P_2=320 \mathrm{~W}
\end{aligned}$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2002
For a cell involving one electron Ecell⊝=0.59 V at 298 K, the equilibrium constant for the cell reaction is:
Given that 2.303 RTF=0.059 V at T=298 K

Options:
A 1.0×102
B 1.0×105
C 1.0×1010
D 1.0×1030

The correct answer is:
1.0×1010

Solution:
Eo=0.59 V
ΔGo=-nFEo; Since n=1
ΔGo=-RT ln K
-1×F×0.59=-RT×2.303 log K
0.59=2.303 RTFlog K
0.59=0.059 log K
10=log K
K=1010

Topic: Chemistry, Electrochemistry, NEET, NEET 2019 (All India)
For a cell, terminal potential difference is $2.2 \mathrm{~V}$ when circuit is open. If it reduces to $1.8 \mathrm{~V}$ when the cell is connected to a resistance of $5 \Omega$. The internal resistance of cell $(r)$ is then:

Options:
A $\frac{10}{9} \Omega$
B $\frac{9}{10} \Omega$
C $\frac{11}{9} \Omega$
D $\frac{5}{9} \Omega$

The correct answer is:
$\frac{10}{9} \Omega$

Solution:
Terminal potential difference
$$
\begin{aligned}
& V=E-I r \\
& \therefore V=E-\left[\frac{E}{R-r}\right] r=\frac{E R}{R+r}
\end{aligned}
$$
From given condition $=E=2.2$ and where $R=5$ then potential $V=1.8 \mathrm{~V}$ Therefore $1.8=\frac{2.2 \times 5}{5+r}$
$\Rightarrow r=\frac{10}{9} \Omega$

Topic: Physics, Current Electricity, NEET, NEET 2002
For a cell, the terminal potential difference is 2.2 V when the circuit is open and is reduced to 1.8 V when the cell is connected to a resistance R=5 Î©. Determine the internal resistance of the cell.

Options:
A 109 Î©
B 910 Ω
C 119 Ω
D 59 Ω

The correct answer is:
109 Î©

Solution:
When the circuit is open, Potential Difference=E=2.2 V
i=2.2r+5

Potential difference across cell =E-ir
1.8=2.2-2.2r+5Ãrâ2.2rr+5=0.4
âr=21.8=109 Î©

Topic: Physics, Current Electricity, NEET
For a certain atom, there are energy levels A, B, C corresponds to energy values EA<EB<EC. Choose the correct option if Î»1, Î»2, Î»3 are the wavelength of radiations corresponding to the transition from C to B, B to A and C to A respectively.

Options:
A λ3=λ1+λ2
B λ3=λ1λ2λ1+λ2
C λ1+λ2+λ3=0
D 3λ2=λ3+2λ2

The correct answer is:
λ3=λ1λ2λ1+λ2

Solution:
E B − E A = hc λ 1
EC-EB=hcλ2
E C − E A = hc λ 3
Or 1λ3=1λ1+1λ2
λ3=λ1λ2λ1+λ2

Topic: Chemistry, Structure of Atom, NEET
For a certain metal Î½=2 Î½0 and the electrons come out with a maximum velocity of 4Ã106 m s-1. If the value of Î½=5 Î½0, then maximum velocity of photoelectrons will be (Î½0 is the threshold frequency)

Options:
A 2Ã107 m s-1
B 8Ã106 m s-1
C 2Ã106 m s-1
D 8Ã105 m s-1

The correct answer is:
8Ã106 m s-1

Solution:
12mv12=2hν0-hν0=hν0 and
12mv22=5hν0-hν0=4 hν0
So, 12mv22=4Ã12mv12
or v2=2v1=2Ã4Ã106=8Ã106 m s-1

Topic: Physics, Dual Nature of Matter, NEET
For a certain reaction a large fraction of molecules has energy more than the threshold energy, yet the rate of reaction is very slow. Why?

Options:
A The reaction is catalyzed by an enzyme.
B Apart from the energy considerations, the colliding molecules should also have proper orientation for effective collision. This condition might not be getting fulfilled in the reaction.
C The reaction is not influenced by the energy of the molecules.
D The reaction is reversible, and the products are continuously converting back into reactants.

The correct answer is:
Apart from the energy considerations, the colliding molecules should also have proper orientation for effective collision. This condition might not be getting fulfilled in the reaction.

Solution:
Correct Option is : (B)
Apart from the energy considerations, the colliding molecules should also have proper orientation for effective collision. This condition might not be getting fulfilled in the reaction.

Topic: Chemistry, Chemical Kinetics, NEET
For a certain reaction, the rate =k [A]2 [B], when the initial concentration of A is tripled keeping concentration of B constant, the initial rate would

Options:
A increase by a factor of three.
B decrease by a factor of nine.
C increase by a factor of six.
D increase by a factor of nine.

The correct answer is:
increase by a factor of nine.

Solution:
The rate law is : rate = \(k[A]^2 \left[B\right]\)...(1)

The rate of the reaction when the concentration of A is Trippled and B is kept constant:

r1=k3A2B

r1=9kA2B.....(2)

From equation first and second:

r1= 9r

Thus, the rate is increased by a factor of 9.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2023 (All India)
For a chemical reaction
$4 \mathrm{~A}+3 \mathrm{~B} \rightarrow 6 \mathrm{C}+9 \mathrm{D}$
rate of formation of $\mathrm{C}$ is $6 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and rate of disappearance of $\mathrm{A}$ is $4 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. The rate of reaction and amount of $B$ consumed in interval of 10 seconds, respectively will be:

Options:
A $1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and $30 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$
B $10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and $10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$
C $1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and $10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$
D $10 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and $30 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$

The correct answer is:
$1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ and $30 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$

Solution:
$$
\begin{aligned}
& \text { Explanation: } 4 \mathrm{~A}+3 \mathrm{~B} \longrightarrow 6 \mathrm{C}+9 \mathrm{D} \\
& \qquad \begin{aligned}
\text { Rate of reaction } & =-\frac{1}{4} \frac{d[\mathrm{~A}]}{d t}=-\frac{1}{3} \frac{d[\mathrm{~B}]}{d t} \\
& =\frac{1}{6} \frac{d[\mathrm{C}]}{d t}=\frac{1}{9} \frac{d[\mathrm{D}]}{d t}
\end{aligned}
\end{aligned}
$$
Rate of reaction $=$ Rate of formation of $\mathrm{C}$.
$$
\begin{aligned}
\therefore \quad r & =\frac{1}{6} \frac{d[\mathrm{C}]}{d t}=\frac{1}{6} \times 6 \times 10^{-2} \\
& =1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
\end{aligned}
$$
Also, rate of reaction $=$ Rate of disappearance of $B$
$$
\begin{aligned}
r & =-\frac{1}{3} \frac{d[\mathrm{~B}]}{d t}=\frac{d[\mathrm{~B}]}{d t}=3 \times r \\
& =3 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
\end{aligned}
$$
Consumption of $B$ in 10 seconds $=3 \times 10^{-2} \times 10$
$$
=30 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}
$$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2022 (Phase 2)
For a chemical reaction,
m1A+m2B→n1C+n2D
The ratio of rate of disappearance of A to that of appearance of C is

Options:
A m1/m2
B m2/n1
C n1/m1
D m1/n1

The correct answer is:
m1/n1

Solution:
For the reaction
m1A+m2B→n1C+n2D
-1m1d[A]dt=+1n1d[C]dt
-d[A]dtd[C]dt=m1n1 .

Topic: Chemistry, Chemical Kinetics, NEET
For a common emitter amplifier, the audio signal voltage across the collector resistance 2 kÎ© is 2 V. If the current amplification factor of the transistor is 200, and the base resistance is 1.5 kÎ©, the input signal voltage and base current is

Options:
A 0.15 V and 10 μA

B 1.015 V and 1 A

C 0.0015 V and 1 mA

D 0.0075 V and 5 μA

The correct answer is:
0.0075 V and 5 μA

Solution:
The collector current
IC=βIb=22×103
IC=1×10-3 A
The base current
Ib=10-3200
Ib=5×10-6 A A
Ib=5 μA and VB=IbRi=5 μA×1.5 kΩ =0.0075 V

Topic: Physics, Semiconductors, NEET
For a cubic crystal structure which one of the following relations indicating the cell characteristics is correct?

Options:
A $a \neq b \neq c$ and $\alpha=\beta=\gamma=90^{\circ}$
B $a=b=c$ and $\alpha \neq \beta \neq \gamma=90^{\circ}$
C $a=b=c$ and $\alpha=\beta=\gamma=90^{\circ}$
D $a \neq b \neq c$ and $\alpha \neq \beta$ and $\gamma \neq 90^{\circ}$

The correct answer is:
$a=b=c$ and $\alpha=\beta=\gamma=90^{\circ}$

Solution:
In a cubic crystal structure
$a=b=c, \alpha=\beta=\gamma=90^{\circ}$

Topic: Chemistry, Solid State, NEET, NEET 2007
For a decomposition of azoisopropane at 270°C it was found that at t = 0, the total pressure was 33.15 mm of Hg and after 3 minutes the total pressure was found to be 46.3 mm of Hg. Calculate the value of K for this reaction.
(CH3)2CHN=NCH(CH2)2→N2+C6H14

Options:
A 0.168 min-1
B 0.173 min-1
C 0.18 min-1
D 0.154 min-1

The correct answer is:
0.168 min-1

Solution:
k=2.303tlogpo(n-1)npo-p1
k=2.3033log33.15(2-1)2×33.15-46.3
k=0.168 min-1

Topic: Chemistry, Chemical Kinetics, NEET
For a diamagnetic substance, susceptibility is

Options:
A small and positive
B small and negative
C large and positive
D large and negative

The correct answer is:
small and negative

Solution:
Susceptibility x of diamagnetic substances is small and negative

χ>0 for paramagnetic
x>>0 for ferromagnetic
χ<0 for diamagnetic

Topic: Physics, Magnetic Properties of Matter, NEET
For a dilute solution containing 2.5 g of a non-volatile non–electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2 o C . Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (in mm of Hg) of the solution is (take K b =0.76 Kkg  mol −1 )

Options:
A 718
B 736
C 724
D 740

The correct answer is:
724

Solution:
ÎTf=iKfm ( i=1 for non-electrolyte)
2=0.76Ãn1001000
n=27.6
Relative lowering in vapour pressure =XB (mole fraction of solute)
ÎPPA0=2/7.6100/18
ÎP760=36760
ÎP=36
P=760-36
P=724 mm of Hg

Topic: Chemistry, Solutions, NEET
For a first order reaction $\mathrm{A} \rightarrow \mathrm{B}$ the reaction rate at reactant concentration of $0.01 \mathrm{M}$ is found to be $2.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$ $\mathrm{s}^{-1}$. The half life period of the reaction is:

Options:
A $30 \mathrm{~s}$
B $220 \mathrm{~s}$
C $300 \mathrm{~s}$
D $347 \mathrm{~s}$

The correct answer is:
$347 \mathrm{~s}$

Solution:
$$
\begin{aligned}
& \text { rate of reaction }=2 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\
& \Rightarrow \text { order of reaction is } n=1 \\
& \text { rate }=\mathrm{K}[\mathrm{A}]^n \\
& {[\mathrm{~A}] }=0.01 \mathrm{M} \\
& \Rightarrow \quad \mathrm{K}=\frac{2 \times 10^{-5}}{0.01}=2 \times 10^{-3} \\
& \mathrm{~K}=\frac{0.693}{t_{1 / 2}} \\
& t_{1 / 2}=\frac{0.693}{2 \times 10^{-3}}=346.5 \mathrm{~s}
\end{aligned}
$$
Related Theory
For a first-order reaction, a graph plotted with $\ln [A]$ on the $Y$-axis and time on the $X$-axis, will be a straight line with a slope of $-k$.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2005
For a first order reaction A→ Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in min-1 is

Options:
A 0.9212
B 0.4606
C 0.2303
D 1.3818

The correct answer is:
0.9212

Solution:
The integrated rate equation for a first order reaction is given by

k=2.303tlogaa-x

Where

k=Rate constanta=initial concentration of the reactantsa-x=concentration of reactants left after time, t.

k=2.3035log0.10.001 =4.6065=0.9212 min-1

Hence, the rate constant of the given reaction is 0.9212 min-1

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2022 (Phase 1)
For a first order reaction,what is the ratio between the time required for 99% completion and the time required for the completion of 90% of reaction.

Options:
A 1
B 6
C 4
D 2

The correct answer is:
2

Solution:
Correct Option is : (D)
2

Topic: Chemistry, Chemical Kinetics, NEET
For a fixed number of moles of a gas, at constant volume, the pressure of the gas increases with increase in temperature due to

Options:
A decreased rate of collision amongst molecules
B Increase in molecular attraction
C Increase in the average molecular speed
D Decrease in mean free path

The correct answer is:
Increase in the average molecular speed

Solution:
When the average speed of the molecule is increased due to an increase in temperature, then the collisions of molecules with the wall of the container also increases which in terms increases of pressure.

Topic: Chemistry, States of Matter, NEET
For a gas, if the ratio of specific heats at constant pressure and constant volume is γ, then the value of degrees of freedom is

Options:
A γ+1γ-1
B γ-1γ+1
C (γ-1)2
D 2γ-1

The correct answer is:
2γ-1

Solution:
Cv=f2RCp=R1+f2Î³ = ratio of specific heats at constant pressure and constant volumeÎ³=CPCvÎ³=1+2n or 2n=Î³-1 or n= 2Î³-1

Topic: Physics, Thermodynamics, NEET
For a gas RCv=0.4, where R is the universal gas constant and Cv is molar specific heat at constant volume. The gas is made up of molecules which are

Options:
A rigid diatomic
B monoatomic
C non- rigid diatomic
D polyatomic

The correct answer is:
rigid diatomic

Solution:
RCv=0.4âCp-CvCv=0.4
CpCv-1=0.4
â´ CpCv=1.4 â Î³=1.4
As Î³=1.4, the gas is diatomic in nature. Example, air molecules.

Topic: Physics, Thermodynamics, NEET
For a gas RCv=0.4, where R is the universal gas constant and Cv is the molar specific heat at constant volume. The gas is made up of molecules which are

Options:
A rigid diatomic
B monoatomic
C non- rigid diatomic
D polyatomic

The correct answer is:
rigid diatomic

Solution:
RCv=0.4âCp-CvCv=0.4
CpCv-1=0.4
â´ CpCv=1.4 â Î³=1.4
As Î³=1.4, the gas is diatomic in nature. Example, air molecules.

Topic: Physics, Thermodynamics, NEET
For a gaseous reaction, following data is given:

A â¶ B, k1=1015 e-2000/T

C â¶ D, k2=1014 e-1000/T

The temperature at which k1=k2 is :

Options:
A 1000 K
B 2000 K
C 868.82 K
D 434.2 K

The correct answer is:
434.2 K

Solution:
When  k1=k2,
1015 e-2000/T=1014 e-1000/T
10=e1000/T
2.303 log 10=1000T
T=434.2 K

Topic: Chemistry, Chemical Kinetics, NEET
For a gaseous reaction, the rate is expressed in terms of dPdt in place of dCdt or dndt, where C is concentration, n is number of moles and ′P′ is pressure of reactant. The three are related as -

Options:
A dPdt=RTVdndt=dCdt
B 1RTdPdt=1Vdndt=dCdt
C dPdt=dndt=dCdt
D None of these

The correct answer is:
1RTdPdt=1Vdndt=dCdt

Solution:
Rate=dCdt
C=nV and C=PRT
dCdt=ddtnV=1V×dndt
ddtPRT=1RT×dPdt

Topic: Chemistry, Chemical Kinetics, NEET
For a gaseous reaction, the rate of reaction is expressed in terms of dPdt instead of dCdt or dndt, where C is concentration and n is the number of moles. Hence, the relation between expressions is

Options:
A dCdt=dndt=VRTdPdt
B dCdt=1Vdndt=1RTdPdt
C dPdt=dndt=dCdt
D none of these

The correct answer is:
dCdt=1Vdndt=1RTdPdt

Solution:
PV=nRT
P=nVRT=CRT
1RTdPdt=1Vdndt=dCdt

Topic: Chemistry, Chemical Kinetics, NEET
For a general reaction A → B, plot of concentration of A vs time is given in the figure below. Answer the following questions on the basis of this graph. (i) What is the order of the reaction? (ii) What is the slope of the curve? (iii) What are the units of rate constant?

Options:
A (i) Zero (ii) +k (iii) mol L–1s–1
B (i) One (ii) –k (iii) mol L–1s–1
C (i) Zero (ii) –k (iii) mol L–1s–1
D (i) Zero (ii) –k (iii) s–1

The correct answer is:
(i) Zero (ii) –k (iii) mol L–1s–1

Solution:
Correct Option is : (C)
(i) Zero (ii) –k (iii) mol L–1s–1

Topic: Chemistry, Chemical Kinetics, NEET
For a given exothermic reaction, Kp and KP' are the equilibrium constants at temperatures T1 and T2 , respectively. Assuming that heat of reaction is constant in temperature range between T1 and T2 , it is readily observed that
T2>T1

Options:
A KP=1KP'
B KP<KP'
C KP=KP'
D KP>KP'

The correct answer is:
KP>KP'

Solution:
log⁡kpkp'=∆H2.303R1T1-1T2
for exothermic reaction
∆H=-ve
T2>T1
In exothermic reactions on increasing temperature value of equilibrium decreases.
So, Kp>Kp'

Topic: Chemistry, Chemical Equilibrium, NEET
For a given exothermic reaction Kp and K'p are the equilibrium constants at temperatures T1 and T2 respectively (T2.>T1 ). Assuming that heat of reaction is constant in temperatures range between T1 and T2, it is readily observed that:

Options:
A Kp>K'p
B Kp<K'p
C Kp=K'p
D Kp=1K'p

The correct answer is:
Kp>K'p

Solution:
log⁡K2K1=∆Ho2.303R1T1-1T2
T2>T1 So Kp> K'p (exothermic reaction)
(Since T2>T1 )

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2014
For a given reaction, energy of activation for forward reaction E a f is 80âkJâ mol -1 and Î H = - 40 kJ mol -1 . A catalyst lowers E a f to 20âkJâ mol â1 . The ratio of energy of activation for reverse reaction before and after addition of catalyst is:

Options:
A 1.0
B 0.5
C 1.2
D 2.0

The correct answer is:
2.0

Solution:
Δ H = E f - E b
- 4 0 = 8 0 - E b
E b =120 kJ/mol,
Catalyst lower the E f to 20 kJ/mol for forward reaction then E f ′ = 20 kJ/mol
We know catalyst decreases the activation energy equal amount in both direction.
E b ′ = 1 2 0 - 6 0 = 60 kJ/mol
E b E b ′ = 1 2 0 6 0 = 2

Topic: Chemistry, Chemical Kinetics, NEET
For a given reaction, ΔH=35.5kJmol-1 and ΔS=83.6Jk-1mol-1 . The reaction is spontaneous at
(Assume that ΔH and ΔS do not vary with tempearature)

Options:
A T > 425 K
B All temperatures
C T > 298 K
D T < 425 K

The correct answer is:
T > 425 K

Solution:
As Both ΔH and ΔS are positive so to keep ΔG=-ve High Temp is favourable so that process become spontaneous.
ΔG=ΔH-T.ΔS
T=ΔHΔS=35.5×100083.6=424.6K
T>425K

Topic: Chemistry, Thermodynamics (C), NEET
For a given reaction, ÎH=35.5 kJ mol-1 and ÎS=83.6 JK-1mol-1. The reaction is spontaneous at : (Assume that ÎH and ÎS do not vary with temperature)

Options:
A T < 425 K
B T > 425 K
C All temperatures
D T > 298 K

The correct answer is:
T > 425 K

Solution:
âµ ÎG=ÎH-TÎS
For a reaction to be spontaneous, ÎG= -ve
i.e., ÎH<TÎS
â´ T>ÎHÎS=35.5Ã103 J83.6 JK-1
i.e., T>425 K

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2017
For a given reaction, ÎH=35.5 kJ mol-1 and ÎS=83.6 JK-1mol-1. The reaction is spontaneous at : (Assume that ÎH and ÎS do not vary with temperature)

Options:
A T < 425 K
B T > 425 K
C All temperatures
D T > 298 K

The correct answer is:
T > 425 K

Solution:
âµ ÎG=ÎH-TÎS
For a reaction to be spontaneous, ÎG= -ve
i.e., ÎH<TÎS
â´ T>ÎHÎS=35.5Ã103 J83.6 JK-1
i.e., T>425 K

Topic: Chemistry, Thermodynamics (C), NEET
For a glass prism (µ = $\sqrt{3}$), the angle of minimum deviation is equal to the angle of the prism. What is the value of the angle of the prism?

Options:
A 60°
B 30°
C 45°
D 90°

The correct answer is:
60°

Solution:
Correct Option is : (A)
60°

Topic: Chemistry, Ray Optics, NEET
For a hypothetical hydrogen like atom, the potential energy of the system is given by U r = - Ke 2 r 3 , where r is the distance between the two particles. If Bohr's model of quantization of angular momentum is applicable then velocity of particle is given by :

Options:
A v = n 2 h 3 Ke 2 8 π 3 m 2
B v = n 3 h 3 8 Ke 2 π 3 m 2
C v = n 3 h 3 2 4 Ke 2 π 3 m 2
D v = n 2 h 3 2 4 Ke 2 π 3 m 2

The correct answer is:
v = n 3 h 3 2 4 Ke 2 π 3 m 2

Solution:
d U r dr = 3 Ke 2 r 4 ⇒ Magnitude of the centripetal force (= Electrostatic force)
∴ 3 Ke 2 r 4 = mv 2 r
and we know mvr = nh 2 π or r = nh 2 π m · v
3 Ke 2 × 8 π 3 m 3 v 3 n 3 h 3 = mv 2 , v = n 3 h 3 2 4 Ke 2 π 3 m 2

Topic: Chemistry, Structure of Atom, NEET
For a hypothetical reaction A(g)+3B(g)â2C(g). ÎH=-100 kJ and ÎS=-200 JK-1 . Then the temperature at which the reaction will be in equilibrium is

Options:
A 500 K
B 480 K
C 520 K
D 510 K

The correct answer is:
500 K

Solution:
At equilibrium ΔG=0 , ΔH-TΔS=0 ∴ T=ΔHΔS

Topic: Chemistry, Thermodynamics (C), NEET
For a linear plot of log(x/m) versus log p in a Freundlich absorption isotherm, which of the following statements is correct? (k and n are constants)

Options:
A log (1/n) appears as the intercept
B Both k and 1/n appear in the slope term
C 1/n appears as the intercept
D Only 1/n appears as the slope

The correct answer is:
Only 1/n appears as the slope

Solution:
According to Freundlich isotherm

xm=k.p1/n
logxm=logk+1nlogP
So intercept is log k and slope is 1n

Topic: Chemistry, Surface Chemistry, NEET
For a material with a refractive index µ ≥ $\sqrt{2}$, light incident at any angle shall be guided along

Options:
A a length perpendicular to the incident face.
B a length parallel to the incident face.
C an angle greater than the critical angle.
D a length at an oblique angle to the incident face.

The correct answer is:
a length perpendicular to the incident face.

Solution:
Correct Option is : (A)
a length perpendicular to the incident face.

Topic: Chemistry, Ray Optics, NEET
For a normal eye, the cornea of eye provides a converging power of $40 \mathrm{D}$ and the least converging power of the eye lens behind the cornea is $20 \mathrm{D}$. Using this information, the distance between the retina and the cornea-eye lens can be estimated to be

Options:
A $5 \mathrm{~cm}$
B $2.5 \mathrm{~cm}$
C $1.67 \mathrm{~cm}$
D $1.5 \mathrm{~cm}$

The correct answer is:
$1.67 \mathrm{~cm}$

Solution:
Given $P_1=40 \mathrm{D}$, and $P_2=20 \mathrm{D}$
We have $P_{\text {eff }}=40 \mathrm{D}+20 \mathrm{D}=60 \mathrm{D}$
$\Rightarrow \quad f=\frac{100}{P_{\text {eff }}}=\frac{100}{60}$
$=1.67 \mathrm{~cm}$

Topic: Physics, Ray Optics, NEET, NEET 2013 (All India)