datavorous/entrance-exam-dataset huggingface dataset:
Assuming very dilute aqueous solution of urea, calculate the vapour pressure of solution (in mm of Hg) of 0.1 moles of urea in 180 grams of water at 25oC is (The vapour pressure of water at 25oC is 24 mm Hg)

Options:
A 2.376
B 20.76
C 23.76
D 24.76

The correct answer is:
23.76

Solution:
From Raoult's law for very dilute solution.
po-pspo=wmÃMW
24-ps24=0.1Ã18180
24-ps=0.24
â´ ps=24-0.24=23.76 mm Hg

Topic: Chemistry, Solutions, NEET
Asthma is characterised by

Options:
A spasm in bronchial muscle
B alveolar wall degradation
C pain in lungs
D damage in diaphragm

The correct answer is:
spasm in bronchial muscle

Solution:
Asthma is a respiratory disorder. It is caused by allergic reactions due to pollen etc. It results in spasm of bronchial muscles and thus the patient experiences shortness of breath and thus difficulty in breathing. The characteristic feature that helps is diagnosis is production of wheezing sound

Topic: Biology, Breathing and Exchange of Gases, NEET
Asthma may be attributed to:

Options:
A Bacterial infection of the lungs
B Allergic reaction of the mast cells in the lungs
C Fungal infection of the lungs
D Accumulation of fluid in the lung

The correct answer is:
Allergic reaction of the mast cells in the lungs

Solution:
Asthma is an allergic reaction characterized by spasm of bronchi muscles because of the effect of histamine released by mast cells.

Occupational asthma is usually caused by inhaling fumes, gases, dust, or other potentially harmful substances during work or job.

Asthma during childhood has impacted millions of children and their families. The majority of children who develop asthma do so before the age of five.

Topic: Biology, Breathing and Exchange of Gases, NEET, NEET 2016 (Phase 1)
Asymptote is a logistic growth curve is obtained when

Options:
A The value of 'r' approaches zero
B K = N
C K > N
D K < N

The correct answer is:
K = N

Solution:
When K = N in a logistics growth curve, it is an asymptote.
It means a population growing in a habitat with limited resources show initially an Iag phase, followed by a phase of acceleration and deceleration and finally an asymptote, i.e. when the population density (N)reaches the carrying capacity (K)
Population growth curve rs logistic,
when responses are limiting the growth, here K is carrying capacity and N is population density.

Topic: Biology, Organisms and Populations, NEET, NEET 2017
At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

Options:
A 38 g
B 60 g
C 45 g
D 70 g

The correct answer is:
70 g

Solution:
Correct Option is : (D)
70 g

Topic: Chemistry, States of Matter, NEET
At $10^{\circ} \mathrm{C}$ the value of the density of a fixed mass of an ideal gas divided by its pressure is $x$. At $110^{\circ} \mathrm{C}$ this ratio is

Options:
A $x$
B $\frac{383}{283} x$
C $\frac{10}{110} x$
D $\frac{283}{383} x$

The correct answer is:
$\frac{283}{383} x$

Solution:
Let the ratio of density and pressure be r.
From Ideal Gas Equation, \(\rho=\frac{\mathrm{PM}}{\mathrm{RT}}\)
\(\begin{aligned}
& \mathrm{r}=\frac{\rho}{\mathrm{P}}=\frac{\mathrm{M}}{\mathrm{RT}} \\
& \therefore \frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{273+10}{273+110}=\frac{283}{383} \\
& \mathrm{r}_2=\frac{283}{383} \quad\left(\text { as, } \mathrm{r}_1=\mathrm{x}\right)
\end{aligned}\)

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2008 (Screening)
At $25^{\circ} \mathrm{C}$ molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is $9.54 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ and at infinite dilution its molar conductance is $238 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is

Options:
A $2.080 \%$
B $20.800 \%$
C $4.008 \%$
D $40.800 \%$

The correct answer is:
$4.008 \%$

Solution:
Given, molar conductance at $0.1 \mathrm{M}$ concentration,
$\lambda_c=9.54 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
Molar conductance at infinite dilution,
$\lambda_c^{\infty}=238 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
We know that, degree of ionisation,
$\begin{aligned} \alpha & =\frac{\lambda_c}{\lambda_c^{\infty}} \times 100 \\ & =\frac{9.54}{238} \times 100=4.008 \%\end{aligned}$

Topic: Chemistry, Electrochemistry, NEET, NEET 2013 (All India)
At $25^{\circ} \mathrm{C}$, the dissociation constant of a base, $\mathrm{BOH}$, is $1.0 \times 10^{-12}$. The concentration of hydroxyl ions in $0.01 \mathrm{M}$ aqueous solution of the base would be:

Options:
A $1.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$
B $1.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$
C $2.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$
D $1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}$

The correct answer is:
$1.0 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}$

Solution:
$$
\begin{gathered}
\mathrm{C}=0.01 \mathrm{M} \\
\mathrm{K}_b=1 \times 10^{-12} \text { at } 25^{\circ} \mathrm{C} \\
\mathrm{BOH} \rightleftharpoons \mathrm{B}^{+}+\mathrm{OH}^{-} \\
\mathrm{C} \quad 0 \quad 0
\end{gathered}
$$
at eq. $\mathrm{C}-\mathrm{C} \alpha \mathrm{C} \alpha \mathrm{C} \alpha$
$$
\begin{aligned}
{\left[\mathrm{OH}^{-}\right] } & =\mathrm{C} \alpha \\
{\left[\mathrm{OH}^{-}\right] } & =\sqrt{\mathrm{K}_b \mathrm{C}}=\sqrt{1 \times 10^{-12} \times 10^{-2}}
\end{aligned}
$$
$$
\left[\mathrm{OH}^{-}\right]=10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}
$$
Related Theory
The activities of the solutes are approximated with molarities. The dissociation constant is an immediate consequence of the law of mass action which describes equilibria in a more general way. The dissociation constant is also sometimes called ionization constant when applied to salts.

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2005
At $100^{\circ} \mathrm{C}$ the $\mathrm{K}_w$ of water is 55 times its value at $25^{\circ} \mathrm{C}$. What will be the $\mathrm{pH}$ of neutral solution? $(\log 55=1.74)$

Options:
A 7.00
B 7.87
C 5.13
D 6.13

The correct answer is:
6.13

Solution:
$$
\begin{aligned}
\mathrm{H}^{+} & =\sqrt{55 \times 10^{-14}} \\
\mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right] \\
\mathrm{pH} & =-\log \left[\sqrt{55 \times 10^{-14}}\right] \\
& =\frac{1}{2}\left[-\log \left(55 \times 10^{-14}\right)\right] \\
& =6.13
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2013 (Karnataka)
At 25° C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm-1cm2mol-1 and at infinite dilution molar conductance is 238 ohm-1cm2mol-1 . The degree of ionisation of ammonium hydroxide at the same concentration and temperature is

Options:
A 20.800%
B 1.008%
C 40.800%
D 2.080%

The correct answer is:
1.008%

Solution:
α=∧m∧m∞=9.54238=0.04008=4.008%

Topic: Chemistry, Electrochemistry, NEET
At 25Â°C the enthalpy change, for the ionization of trichloroacetic acid is +6.3 kJ mol-1 and the entropy change, is +0.0084 kJ mol-1 K-1. Then pKa of tricholoro acetic acid is

Options:
A 1.74
B 2.52
C 0.66
D 4.72

The correct answer is:
0.66

Solution:
ÎGÂ°=ÎHÂ°-TÎSÂ°
ÎGÂ°=6.3-(298Ã0.0084)=3.8 KJ.mol-1
∴   Δ G o =2.303 RTlnK
logKeq=logKa=-pKa=-ÎGÂ°2.303RT
-pKa=-ÎGÂ°2.303RT=3.82.303Ã0.008314Ã298
=0.66

Topic: Chemistry, Thermodynamics (C), NEET
At 25°C and 760 mm of Hg pressure, a gas occupies 600 mL volume. What will be its pressure at a height where the temperature is 10°C and the volume of the gas is 640 mL?

Options:
A 654.8 mm Hg
B 676.6 mm Hg
C 625.4 mm Hg
D 698.1 mm Hg

The correct answer is:
676.6 mm Hg

Solution:
Correct Option is : (B)
676.6 mm Hg

Topic: Chemistry, States of Matter, NEET
At 27 o C, H 2 gas is leaked through a tiny hole into a vessel for 20 minutes. Another unknown gas, at the same temperature and pressure, as that of H 2 , is leaked through the same hole for 20 minutes. After the effusion of the gases the mixture exerts a pressure of 6 atm. If H 2 is 0.7 mol and volume of vessel is 3 litre, calculate molecular weight of the unknown gas.

Options:
A 1033
B 2032
C 1325
D 2132

The correct answer is:
1033

Solution:
Mole of gaseous mixture,
n H 2  + Unknown gas  =  PV RT
= 6×3 0.0821×300 =0.7308 mole
n (Unknown gas) = 0.7308−0.7 = 0.0308 mole
Mole of H 2 leaked Mole of unknown gas leaked  =  M Unknown M H 2
0.7 0.0308  =  M 2
M =  1033 g mol −1

Topic: Chemistry, States of Matter, NEET
At 35oC the vapour pressure of CS2 is 512 mm Hg, and of acetone, CH3COCH3 , is 344 mm Hg. A solution of CS2 and acetone in which the mole fraction of CS2 is 0.25 has a total vapour pressure of 600 mm Hg. Which of the following statements about solution of acetone - CS2 is true ?

Options:
A A mixture of 100 ml of acetone and 100 ml of CS2 has a total volume of 200 ml.
B When acetone and CS2 are mixed at 35oC , heat must be absorbed in order to produce a solution at 35oC .
C When acetone and CS2 are mixed at 35oC , heat is released.
D Raoult's law is obeyed by both CS2 and acetone for the solution in which the mole fraction of CS2 is 0.25.

The correct answer is:
When acetone and CS2 are mixed at 35oC , heat must be absorbed in order to produce a solution at 35oC .

Solution:
PT=PAÂ°XA+PBÂ°XB=512Ã0.25+344Ã0.75=386
â´ The solution is showing positive deviation
So For a solution of
Acetone+CS2; ÎH=+ve

Topic: Chemistry, Solutions, NEET
At 35ºC, the vapour pressure of CS2 is 512 mm Hg and that of acetone is 344 mm Hg. A solution of CS2 in acetone has a total vapour pressure of 600 mm Hg. The false statement amongst the following is

Options:
A A mixture of 100 mL CS2 and 100 mL acetone has a volume <200 mL
B Roult's law is not obeyed by this system
C Heat must be absorbed in order to produce the solution at 35ºC
D CS2 and acetone are less attracted to each other than to themselves

The correct answer is:
A mixture of 100 mL CS2 and 100 mL acetone has a volume <200 mL

Solution:
The vapour pressure of mixture (= 600 mm Hg) is greater than the individual vapour pressure of its constituents (Vapour pressure of CS2= 512 mm Hg, acetone = 344 mm Hg).
Hence, the solution formed shows positive deviation from Raoult's law.
1 Δsol.H>0,
2 Raoult's law is not obeyed
3 Δsol. Volume > 0
4 CS2 and Acetone are less attracted to each other than to themselves.

Topic: Chemistry, Solutions, NEET
At 37°C , the osmotic pressure of blood is 8.21 atm. The amount of glucose that should be used per litre for an intravenous injection so that it becomes isotonic with blood is

Options:
A 117 g
B 58.06 g
C 108 g
D 580.6 g

The correct answer is:
58.06 g

Solution:
C=ÏRT=8.210.0821Ã310
W180Ã1=8.210.0821Ã310
W =  8.21 × 180 0.0821 × 310  = 58.06 g

Topic: Chemistry, Solutions, NEET
At 80âC , the vapour pressure of pure liquid 'A' is 520âmmââHg and that of pure liquid 'B' is 1000âmmââHg. If a mixture solution of 'A' and 'B' boils at 80âC and 1 atm pressure, the amount of 'A' in the mixture is:
( 1âatmâ=760âmmââHg )

Options:
A 52 mole percent
B 34 mole percent
C 48 mole percent
D 50 mole percent

The correct answer is:
50 mole percent

Solution:
PT = PA∘XA+PB∘XB
Mixture solution boil at 1 atm = 760 mm Hg = total pressure of solution.
7 6 0 = 520X A + 1 0 00 1 - X A
⇒ 7 6 0 = 5 2 0 X A + 1 0 0 0 - 1 0 0 0 X A ⇒ 4 8 0 X A = 2 4 0 ​
X A =0.5,
mole percentage of A=50%

Topic: Chemistry, Solutions, NEET
At 90°C , pure water has [H3O+]=10-6 mol litre-1 . What is the value of Kw at 90° C?

Options:
A 10-6
B 10-12
C 10-14
D 10-8

The correct answer is:
10-12

Solution:
At 90 o C,â H 3 O + â=â OH â
= 10 −6  M
So Kw=[H3O+][OH-]
= 10 â6 âÃ 10 â6
= 10 â12

Topic: Chemistry, Ionic Equilibrium, NEET
At 100oC the vapour pressure of a solution of 6.5 g of a non-volatile solute in 100 g water is 732 mm. If Kb=0.52 , the boiling point of this solution will be:

Options:
A 101oC
B 100oC
C 102oC
D 103oC

The correct answer is:
101oC

Solution:
At 100oC (boiling point)
Vapour pressure of water Po=Patm=760 mm
∴Po-PsPo=Xsolute
⇒760-732760=nsolutensolvent{m=molar mass of solute}
⇒28732=6.5/m100/18
⇒m=6.5×18×73228×100≈30.6
Now,
∆Tb=Kb× molality
=0.52×6.5/30.60.1=1.1°C
∴ Boiling point of solution =100+1.1=101.1oC ≃101°C

Topic: Chemistry, Solutions, NEET, NEET 2016 (Phase 1)
At \(293 \mathrm{~K}\), if one starts with \(1.00 \mathrm{~mol}\) of acetic acid and \(0.18 \mathrm{~mol}\) of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

Options:
A $K_c=0.157$
B $K_c=0.171$
C $K_c=1.000$
D $K_c=1.156$

The correct answer is:
$K_c=0.171$

Solution:
Correct Option is : (B)
$K_c=0.171$

Topic: Chemistry, Ionic Equilibrium, NEET
At 298 K , Kpfor the reaction N2O4(g) ⇌ 2NO2(g) is 0.98. Predict whether the reaction is spontaneous or not.

Options:
A It is a spontaneous process
B It is a non-spontaneous process
C Whether it is a spontaneous or non-spontaneous process depends on the situation
D More information is required for this judgement.

The correct answer is:
It is a spontaneous process

Solution:
Correct Option is : (A)
It is a spontaneous process

Topic: Chemistry, Thermodynamics (C), NEET
At 298 K the molar conductivities at infinite dilution â§mo of NH4Cl, KOH and KCl are 152.8, 272.6 and 149.8 S cm2 mol-1 respectively. The â§mo of NH4OH in S cm2 mol-1 and % dissociation of 0.01 M NH4 OH with â§m=25.1 S cm2 mol-1 at the same temperature are

Options:
A 275.6, 0.91
B 275.6, 9.1
C 266.6, 9.6
D 30, 84

The correct answer is:
275.6, 9.1

Solution:
â§moNH4OH= â§moNH4Cl+KOH- â§moKCl
=152.8+272.6-149.8
=275.6 S cm2 mol-1
Degree of dissociation Î±=â§mâ§mo=25.1275.6
=0.091
â´ % degree of dissociation = 9.1%

Topic: Chemistry, Electrochemistry, NEET
At 298 K, what is the physical state of alkanes containing 18 carbon atoms or more?

Options:
A Gaseous
B Liquid
C Solid
D Variable

The correct answer is:
Solid

Solution:
Alkanes containing 18 carbon atoms or more are in the solid state at 298 K due to their larger molecular size and increased intermolecular van der Waals forces.

Topic: Chemistry, Hydrocarbons, NEET
At 300° C the pressure necessary to obtained 80% dissociation of PCl5 is numerically equal to

Options:
A KP=17.7p
B KP=1.77p
C K P = 2.22p
D KP=177p

The correct answer is:
K P = 2.22p

Solution:
PCl511-0.8⇌PCl500.8+Cl200.8
Total number of moles at equilibrium =0.2+0.8+0.8=1.8
KP=(PPCl3)(PCl2)(PPCl5) = 0.8 1.8 p × 0.8 1.8 p 0.2 1.8 p =2.22p

Topic: Chemistry, Chemical Equilibrium, NEET
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

Options:
A 0.099 M
B 0.223 M
C 0.125 M
D 0.061 M

The correct answer is:
0.061 M

Solution:
Correct Option is : (D)
0.061 M

Topic: Chemistry, Solutions, NEET
At 300 K, half life of a gaseous reactant initially at 58 KPa is 320 min. When the pressure is 29 KPa, the half life is 160 mm. The order of the reaction is

Options:
A 1
B 2
C 3
D 0

The correct answer is:
0

Solution:
t1/2â1an-1
t2t1=a1a2n-1
160320=5829n-1
n = 0.

Topic: Chemistry, Chemical Kinetics, NEET
At 450K, Kp is 2.0 × 10¹⁰ per bar for the given reaction at equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) What is Kc at this temperature?

Options:
A 7.48 × 10-15
B 5.23 × 10⁻¹¹
C 4.35 × 10⁻¹⁶
D 5.23 × 10¹¹

The correct answer is:
7.48 × 10-15

Solution:
Correct Option is : (A)
7.48 × 10-15

Topic: Chemistry, Ionic Equilibrium, NEET
At 550âK , the K c for the following reaction is 10 4 mol â1 âL

X(g)â+âY(g)ââZ(g)

At equilibrium, it was observed that

X â=â 1 2 â Y â=â 1 2 â Z

What is the value of Z in mol L â1 at equilibrium?

Options:
A 2×10-4
B 10-4
C 2×104
D 104

The correct answer is:
2×10-4

Solution:
X(g) + Y(g)⇌Z(g)
K c â=â Z X Y or 10 4 â=â Z X Y
â´â Z â=â 10 4 X â Y
=â 10 4 âÃâ 1 2 â[Z][Z]
(Given, [X]â=â 1 2 â[Y]â=â 1 2 [Z] )
â´â[Z]â=â2âÃâ 10 â4 mol L â1

Topic: Chemistry, Chemical Equilibrium, NEET
At 673 K, which of the following is the more effective reducing agent: C (Carbon) or CO (Carbon Monoxide)?

Options:
A Carbon [C]
B Carbon Monoxide [CO]
C Both are equally effective
D Neither is effective as a reducing agent at this temperature

The correct answer is:
Carbon Monoxide [CO]

Solution:
Correct Option is : (B)
Carbon Monoxide [CO]

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
At 700 K, the equilibrium constant for the reaction H2(g) + I2(g) ⇌ 2HI(g) is 54.8. If 0.5 mol L-1 of HI(g) is present at equilibrium at 700 K, what are the concentrations of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?

Options:
A 0.25 mol L-1for H2and 0.25 mol L-1for I2
B 0.35 mol L-1for H2and 0.35 mol L-1for I2
C 0.068 mol L-1for H2and 0.068 mol L-1for I2
D 0.30 mol L-1for H2and 0.20 mol L-1for I2

The correct answer is:
0.068 mol L-1for H2and 0.068 mol L-1for I2

Solution:
Correct Option is : (C)
0.068 mol L-1for H2and 0.068 mol L-1for I2

Topic: Chemistry, Ionic Equilibrium, NEET
At \(1024 \mathrm{~K}\), the equilibrium constant \(K_c\) for the reaction \(\mathrm{H}_2(g)+\mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{HBr}(g)\) is \(1.6 \times 10^{\wedge} 5\). If 10.0 bar of \(\mathrm{HBr}\) is introduced into a sealed container at \(1024 \mathrm{~K}\), what will be the equilibrium pressures of \(\mathrm{H}_2, \mathrm{Br}_2\), and \(\mathrm{HBr}\) ?

Options:
A \(4.0 \times 10^5\) bar, \(5.0 \times 10^5\) bar, \(1.0 \times 10^5\) bar
B \(1.0 \times 10^5\) bar, \(2.5 \times 10^5\) bar, \(2.5 \times 10^5\) bar
C \(3.3 \times 10^5\) bar, \(3.3 \times 10^5\) bar, \(3.3 \times 10^5\) bar
D \(5.0 \times 10^5\) bar, \(4.0 \times 10^5\) bar, \(1.0 \times 10^5\) bar

The correct answer is:
\(1.0 \times 10^5\) bar, \(2.5 \times 10^5\) bar, \(2.5 \times 10^5\) bar

Solution:
Correct Option is : (B)
\(1.0 \times 10^5\) bar, \(2.5 \times 10^5\) bar, \(2.5 \times 10^5\) bar

Topic: Chemistry, Ionic Equilibrium, NEET
At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2in equilibrium with solid carbon has 90.55% CO by mass. C(s) + CO2(g) ⇌ 2CO(g) Calculate Kc for this reaction at the above temperature.

Options:
A 3.6 × 107
B 8.0 × 106
C 0.154
D 6.0 × 107

The correct answer is:
0.154

Solution:
Correct Option is : (C)
0.154

Topic: Chemistry, Ionic Equilibrium, NEET
At 3000 K, the equilibrium partial pressure of CO2,CO and O2 are 0.6, 0.4 and 0.2 atm respectively. Kp for the reaction, 2CO2â2CO+O2 is

Options:
A 0.088
B 0.0533
C 0.133
D 0.177

The correct answer is:
0.088

Solution:
2CO2 â 2CO+O2
At equili. 0.6 atm 0.4 atm 0.2 atm
â´ Kp=pCO2ÃpO2pCO22
=0.42(0.2)0.62=0.088

Topic: Chemistry, Chemical Equilibrium, NEET
At a certain place, the angle of dip is 60Â° and the horizontal component of the earth's magnetic field BH is 0.8Ã10-4 T. The earth's overall magnetic field is

Options:
A 1.5×10-4 T
B 1.6×10-3 T
C 1.5×10-3 T
D 1.6×10-4 T

The correct answer is:
1.6×10-4 T

Solution:
Given,
BH=0.8×10-4T
θ=60o
Be= ?
We know that,
BH=Becos⁡θ
0.8×10-4=Becos⁡60o
Be=0.8×10-412
=1.6×10-4 T

Topic: Physics, Magnetic Properties of Matter, NEET
At a certain temperature the equilibrium constant KC is 0.25 for the reaction
A(g)+B(g)⇌C(g)+D(g)
If we take 1 mole of each of four gases in a 10 litre container, what would be the equilibrium concentration of A(g)?

Options:
A 0.331 M
B 0.033 M
C 0.133 M
D 1.33 M

The correct answer is:
0.133 M

Solution:
The give reaction will go in backward direction
Kc=(A)×(B)(C)×(D)
4=(1+α)2(1-α)2
2=1+α1-α
2-2α=1+α
α=13=0.33
[A]=1+α10
=1+0.3310=0.133 M

Topic: Chemistry, Chemical Equilibrium, NEET
At a concentration of 10 â3 M, a detergent C12H25SO4-Na+ solution becomes a colloidal solution. On an average 1013 colloidal particles are present in 1 mm 3. Calculate the average number of ions contained in one colloidal particle (micelle)? [Given N A =6Ã 10 23]

Options:
A 6 × 107
B 10
C 60
D None of these

The correct answer is:
60

Solution:
No. of sodium lauryl sulphate CH3CH211SO4-Na+ in 1 litre solution
= 10 â3 Ã6Ã10 23
=6Ã10 20
No. of sodium lauryl sulphate per mm 3 =6Ã 10 14
No. of colloidal particles per mm 3 = 10 13
No. of molecules per colloidal particle
= 6 Ã 1 0 1 4 1 0 1 3
=60

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
At a constant temperature, which of the following aqueous solutions will have the maximum vapour pressure?
(Mol.âwt.âNaCl = 58.5,â H 2 SO 4 â=â98 .0 g mol â1 )

Options:
A 1 molal NaCl (aq)
B 1 molar NaCl (aq)
C 1 molal  H 2 SO 4 aq
D 1 molar  H 2 SO 4 aq

The correct answer is:
1 molal NaCl (aq)

Solution:
One molar (1 M) aqueous solution is more concentrated than one molal aqueous solution of the same solute. In solution, H2SO4 provides three ions. While NaCl provides two ions. Hence, vapour pressure of solution of NaCl is higher (as it gives less ions). Therefore, 1 molal NaCl will have the maximum vapour pressure.

Topic: Chemistry, Solutions, NEET
At a distance 320 km above the surface of the earth, the value of acceleration due to gravity will be lower than its value on the surface of the earth by nearly (radius of earth = 6400 km)

Options:
A 2%
B 6%
C 10%
D 14%

The correct answer is:
10%

Solution:
g′g=1-2hR=1-2×3206400=1-110=910
∴ % decrease in g=g-g′g×100
=1100×100=10%

Topic: Physics, Gravitation, NEET
At a given temperature and pressure, the equilibrium constant values for the equilibria are given below:
$3 \mathrm{~A}_2+\mathrm{B}_2 \rightleftharpoons 2 \mathrm{~A}_3 \mathrm{~B}, \mathrm{~K}_1$
$\mathrm{A}_3 \mathrm{~B} \rightleftharpoons \frac{3}{2} \mathrm{~A}_2+\frac{1}{2} \mathrm{~B}_2, \mathrm{~K}_2$
The relation between $\mathrm{K}_1$ and $\mathrm{K}_2$ is :

Options:
A $\mathrm{K}_1^2=2 \mathrm{~K}_2$
B $\mathrm{K}_2=\frac{\mathrm{K}_1}{2}$
C $\mathrm{K}_1=\frac{1}{\sqrt{\mathrm{K}_2}}$
D $\mathrm{K}_2=\frac{1}{\sqrt{\mathrm{K}_1}}$

The correct answer is:
$\mathrm{K}_2=\frac{1}{\sqrt{\mathrm{K}_1}}$

Solution:
$3 \mathrm{~A}_2+\mathrm{B}_2 \rightleftharpoons 2 \mathrm{~A}_3 \mathrm{~B} ; \mathrm{K}_1$
$2 \mathrm{~A}_3 \mathrm{~B} \rightleftharpoons 3 \mathrm{~A}_2+\mathrm{B}_2 ; \mathrm{K}^{\prime}=\frac{1}{\mathrm{~K}_1}$
$\frac{1}{2} A_3 B \rightleftharpoons \frac{3}{2} A_2+\frac{1}{2} B_2 ; K_2=\sqrt{K^{\prime}}$
$\mathrm{K}_2=\sqrt{\mathrm{K}^{\prime}}$
$=\frac{1}{\sqrt{K_1}}$

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2024 (Re-NEET)
At a particular instant, a stationary observer on the ground sees a package falling with a speed v1 at an angle Î¸ to the vertical. To a pilot flying horizontally with a constant speed v relative to the ground, the package appears to be falling vertically with a speed v2 ( at that same instant ). The speed of the pilot relative to the ground (v) is

Options:
A v 1 2 + v 2 2 1 / 2
B v 1 - v 2 v 2 - v 1
C v 1 2 - v 2 2 1 / 2
D v 1 - v 2

The correct answer is:
v 1 2 - v 2 2 1 / 2

Solution:
we can see from the diagram, vector v1 is resultant of two vectors v and v2

Hence v2+v22=v12
âv=v12-v22

Topic: Physics, Motion In Two Dimensions, NEET
At a particular locus, frequency of $A$ allele is 0.6 and that of a is 0.4 . What would be the frequency of heterozygotes in a random mating population at equilibrium?

Options:
A 0.36
B 0.16
C 0.24
D 0.48

The correct answer is:
0.48

Solution:
In a stable population, for a gene with two alleles, 'A' (dominant) and 'a' (recessive),
the frequency of ' $\mathrm{A}^{\prime}=\mathrm{p}$
the frequency of ' $a$ ' $=q$,
According Hardy-Weinberg equation:
$$
\mathrm{p}^2+2 p q+q^2=1
$$
where $\mathrm{p}^2=$ Frequency of $\mathrm{A}$ (homozygous dominant) individuals $\mathrm{q}^2=$ Frequency of aa (homozygous recessive) individuals
$2 \mathrm{pq}=$ Frequency of $\mathrm{Aa}$ (heterozygous) individuals.
$$
\begin{aligned}
& \therefore 2 \mathrm{pq}(\text { frequency of heterozygote })= \\
& 2 \times 0.6 \times 0.4=0.48
\end{aligned}
$$

Topic: Biology, Evolution, NEET, NEET 2005