datavorous/entrance-exam-dataset huggingface dataset:
At $10^{\circ} \mathrm{C}$ the value of the density of a fixed mass of an ideal gas divided by its pressure is $\mathrm{x}$. At $110^{\circ} \mathrm{C}$ this ratio is

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

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

Solution:
$$
\begin{aligned}
P V & =n R T \\
& \Rightarrow P \cdot \frac{m}{\rho}=\frac{m}{M} R T \\
& \Rightarrow \frac{\left(\frac{\rho}{P}\right)_f}{\left(\frac{\rho}{P}\right)_i}=\frac{T_i}{T_f}=\frac{10+273}{110+273}=\frac{283}{383} \\
& \Rightarrow \frac{\left(\frac{\rho}{P}\right)_f}{x}=\frac{283}{383} \\
& \therefore\left(\frac{\rho}{P}\right)_f=\frac{283}{383} x
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2008 (Mains)
At $25^{\circ} \mathrm{C}$ and 1 bar which of the following has a non-zero $\Delta \mathrm{H}_{\mathrm{f}}^{\circ} ?$

Options:
A $\operatorname{Br}_{2}(\ell)$
B C (graphite)
C $\mathrm{O}_{3}(\mathrm{~g})$
D $\mathrm{I}_{2}(\mathrm{~s})$

The correct answer is:
$\mathrm{O}_{3}(\mathrm{~g})$

Solution:
Ozone is allotropic form of oxygen and is of higher energy (by $68 \mathrm{~K}$ Cal mol $^{-1}$ ) than $\mathrm{O}_{2}$. Hence it can not be taken as the reference in standard state.

Topic: Chemistry, Thermodynamics (C), JEE Main
At $25^{\circ} \mathrm{C}, \mathrm{pH}$ of a $10^{-8} \mathrm{M}$ aqueous $\mathrm{KOH}$
solution will be

Options:
A 6
B 7.02
C 8.02
D 9.02

The correct answer is:
7.02

Solution:
$\mathrm{NaOH} \longrightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-} \quad\left[\mathrm{OH}^{-}\right]=10^{-8} \mathrm{M}$
$\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}$ $\left[\mathrm{OH}^{-}\right]=10^{-7} \mathrm{M}$
$$
\begin{aligned}
\therefore\left[\mathrm{OH}^{-}\right]_{\text {total }} &=\left(10^{-8}+10^{-7}\right) \mathrm{M} \\
&=10^{-7}(1.1) \\
&=1.1 \times 10^{-7} \\
\therefore \quad \mathrm{pOH} &=\log 1.1 \times 10^{-7}=6.98 \\
\therefore \quad \mathrm{pH} &=14-6.98 \\
&=7.02
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main
At $25^{\circ} \mathrm{C}$, the molar conductances at infinite dilution for the strong electrolytes $\mathrm{NaOH}, \mathrm{NaCl}$ and $\mathrm{BaCl}_2$ are $248 \times 10^{-4}, 126 \times 10^{-4}$ and $280 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1}$ respectively, $\lambda_m^{\circ} \mathrm{Ba}(\mathrm{OH})_2$ in $\mathrm{Sm}^2 \mathrm{~mol}^{-1}$ is

Options:
A $52.4 \times 10^{-4}$
B $524 \times 10^{-4}$
C $402 \times 10^{-4}$
D $262 \times 10^{-4}$

The correct answer is:
$524 \times 10^{-4}$

Solution:
$\begin{aligned} & \mathrm{BaCl}_2+2 \mathrm{NaOH} \longrightarrow \mathrm{Ba}(\mathrm{OH})_2+2 \mathrm{NaCl} \\ & \lambda_{m \mathrm{Ba}(\mathrm{OH})_2}^{\infty}=\lambda_{m \mathrm{BaCl}_2}^{\infty}+2 \lambda_{m \mathrm{NaOH}}^{\infty}-2 \lambda_{m \mathrm{NaCl}}^{\infty}\end{aligned}$
$\begin{aligned} & =280 \times 10^{-4}+2 \times 248 \times 10^{-4} \\ & \quad-2 \times 126 \times 10^{-4} \\ & =(280+496-252) \times 10^{-4} \\ & =524 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1}\end{aligned}$

Topic: Chemistry, Electrochemistry, JEE Main
At $25^{\circ} \mathrm{C}$, the solubility product of $\mathrm{Mg}(\mathrm{OH})_2$ is $1.0 \times 10^{-11}$. At which $\mathrm{pH}$, will $\mathrm{Mg}^{2+}$ ions start precipitating in the form of $\mathrm{Mg}(\mathrm{OH})_2$ from a solution of $0.001 \mathrm{M} \mathrm{Mg}^{2+}$ ions ?

Options:
A
9

B
10

C
11

D
8

The correct answer is:
10

Solution:
$\mathrm{Mg}^{2+}+2 \mathrm{OH}^{-} \rightleftharpoons \mathrm{Mg}(\mathrm{OH})_2$
$\mathrm{~K}_{\mathrm{sp}}=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2$
$\left[\mathrm{OH}^{-}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{sp}}}{\left[\mathrm{Mg}^{2+}\right]}}=10^{-4}$
$\therefore \mathrm{p}^{\mathrm{OH}}=4$ and $\mathrm{p}^{\mathrm{H}}=10$

Topic: Chemistry, Ionic Equilibrium, JEE Main
At $25^{\circ} \mathrm{C}$, the solubility product of $\mathrm{Mg}(\mathrm{OH})_2$ is $1.0 \times 10^{-11}$. At which $\mathrm{pH}$, will $\mathrm{Mg}^{2+}$ ions start precipitating in the form of $\mathrm{Mg}(\mathrm{OH})_2$ from a solution of $0.001 \mathrm{M} \mathrm{Mg}^{2+}$ ions ?

Options:
A
9

B
10

C
11

D
8

The correct answer is:
10

Solution:
$\mathrm{Mg}^{2+}+2 \mathrm{OH}^{-} \rightleftharpoons \mathrm{Mg}(\mathrm{OH})_2$
$\mathrm{~K}_{\mathrm{sp}}=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2$
$\left[\mathrm{OH}^{-}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{sp}}}{\left[\mathrm{Mg}^{2+}\right]}}=10^{-4}$
$\therefore \mathrm{p}^{\mathrm{OH}}=4$ and $\mathrm{p}^{\mathrm{H}}=10$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2010
At $80^{\circ} \mathrm{C}$, the vapour pressure of pure liquid ' $A$ ' is $520 \mathrm{~mm} \mathrm{Hg}$ and that of pure liquid ' $B$ ' is $1000 \mathrm{~mm}$ $\mathrm{Hg}$. If a mixture solution of ' $A$ ' and ' $B$ ' boils at $80^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure, the amount of ' $A$ ' in the mixture is $(1 \mathrm{~atm}=760 \mathrm{~mm} \mathrm{Hg})$

Options:
A
52 mol percent

B
34 mol percent

C
48 mol percent

D
50 mol percent

The correct answer is:
50 mol percent

Solution:
$$
\begin{aligned}
& P_T=P_A^{\circ} X_A+P_B^{\circ} X_B \\
& 760=520 X_A+P_B^{\circ}\left(1-X_A\right) \\
& \Rightarrow X_A=0.5
\end{aligned}
$$
Thus, mole $\%$ of $A=50 \%$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2008
At 20°C, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20°C above an equimolar mixture of benzene and methyl benzene is __×10-2. (Nearest integer)

Options:

The correct answer is:
78

Solution:
PB∘=70PT∘=20XB=0.5=XT

Now yB=XBPB∘XBPBo+XTPT∘

=70×0.570×0.5+20×0.5=78×10-2

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (20 Jul Shift 1)
At 25°C,50 g of iron reacts with HCl to form FeCl2. The evolved hydrogen gas expands against a constant pressure of 1 bar. The work done by the gas during this expansion is -_______ J.

(Round off to the Nearest Integer)

[Given : R=8.314 J mol-1 K-1. Assume, hydrogen is an ideal gas]

[Atomic mass off Fe is 55.85 u]

Options:

The correct answer is:
2218

Solution:
T=298 K, R=8.314 JmolK

→ Chemical reaction is

Fe50g+2HCl→FeCl2P=1 bar+H2g

=5055.85 mol

→ Work done for 1 mol gas

=-Pext×ΔV

=ΔngRT

=-1×8.314×298 J

→ Work done for 5055.85 mol of gas

=-8.314×298×5055.85 J

=-2218.059 J

≃-2218 J

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (16 Mar Shift 2)
At 25 °C and 1 atm pressure, the enthalpy of combustion of benzene 1 and acetylene g are -3268 kJ mol-1 and -1300 kJ mol-1, respectively. The change in enthalpy for the reaction 3C2H2g→C6H6l, is

Options:
A +324 kJ mol-1
B +632 kJ mol-1
C -632 kJ mol-1
D -732 kJ mol-1

The correct answer is:
-632 kJ mol-1

Solution:
Standard enthalpy of combustion is defined as the enthalpy change when one mole of a compound is completely burnt in oxygen with all the reactants and products in their standard state under standard conditions (298K and 1 bar pressure)

ΔHr=ΔHCreactant-ΔHCproduct

=3ΔHccC2H2,g-ΔHcC6H6,l

=3×-1300--3268

=-3900+3268=-632 kJ/mole

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2022 (25 Jun Shift 2)
At 25°C, the enthalpy of the following processes are given:

H2( g)+O2( g)→2OH(g)ΔHo=78 kJ mol-1

H2( g)+1/2O2( g)→H2O(g)ΔH0=-242 kJ mol-1

H2( g)→2H(g)ΔHo=436 kJ mol-1

1/2O2( g)→O(g)ΔH0=249 kJ mol-1

What would be the value of X for the following reaction? (Nearest integer)

H2Og→Hg+OHg ∆Ho=X kJmol-1

Options:

The correct answer is:
499

Solution:
2H2O(g)→2H2( g)+O2( g) +(242×2)kJmol-1 ....(1)

H2( g)+O2( g)→2OH +78 kJ mol-1 ....(2)

H2( g)→2H(g) +436 kJ mol-1 ....(3)
Using Hess's law of constant heat summation, adding equations(1), (2) and (3) and dividing the net result by 2 we get:

⇒H2O(g)→H(g)+OH(g) 998×12=+499 kJ mol-1
Thus X = 499

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (01 Feb Shift 1)
At 25Â°C the pH of a 10â8 M HCl solution in water is

Options:
A 8
B –8
C between 7 and 8
D between 6 and 7

The correct answer is:
between 6 and 7

Solution:
pH of 10â8 M HCl will be lesser than 7 as it is acidic solution.
H + = H + HCl + H + H 2 O
H + = 10 −8 + 10 −7
∴  H + =1.1× 10 −7
pH = 6.96

Topic: Chemistry, Ionic Equilibrium, JEE Main
At 25 ° C the pH of water is 7. When temperature of water is increased to 70 ° C than pH of water and nature of water is

Options:
A pH will decrease and the sample becomes acidic
B pH will increase but the sample will remain neutral
C pH will remain constant as 7.
D pH will decrease but the sample will remain neutral.

The correct answer is:
pH will decrease but the sample will remain neutral.

Solution:
High temperature pH and pOH of water decreases so, it will be less then 7 but H + = OH − so solution become neutral.

Topic: Chemistry, Ionic Equilibrium, JEE Main
At 25 ° C the pH of water is 7. When temperature of water is increased to 70 ° C than pH of water and nature of water is

Options:
A pH will decrease and the sample becomes acidic
B pH will increase but the sample will remain neutral
C pH will remain constant as 7.
D pH will decrease but the sample will remain neutral.

The correct answer is:
pH will decrease but the sample will remain neutral.

Solution:
High temperature pH and pOH of water decreases so, it will be less then 7 but H + = OH − so solution become neutral.

Topic: Chemistry, Ionic Equilibrium, NEET
At 25oC , nitrogen exists as N2 and phosphorous exists as P4 because

Options:
A N2 has valence electrons only in bonding and nonbonding orbitals, while P has valence electrons in both bonding and antibonding orbitals
B higher electronegativity of N favours formation of multiple bonds
C bigger size of P does not favour multiple bonds
D P has preference to adapt structures with small bond angles

The correct answer is:
bigger size of P does not favour multiple bonds

Solution:
Due to larger atomic size P is unable to form pi bond and so it is tetra-atomic P4 in which each P atom is linked with 3 other P atoms by 3 sigma bonds. But, due to smaller atomic size N forms 1 sigma and 2 pi bonds i.e. triple bonds with other N atom and exists as diatomic molecule N2 .

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
At 25oC, pH range of phenolphthalein is 8-10. At 100oC, pH range of phenolphthalein would be -

Options:
A pH range remain unaffected by the temperature
B pH range is altered to 7 - 9
C pH range is altered to 7 - 11
D pH range is altered to 8 - 11

The correct answer is:
pH range is altered to 7 - 9

Solution:
The pH range of the indicator is pKIn±1. On increase in temperature pKIn decreases therefore pH range of indicator decreases.

Topic: Chemistry, Ionic Equilibrium, JEE Main
At 27°C, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is g mol-1 (Nearest integer)
(Given : R=0.083 L bar -1mol-1 )

Options:

The correct answer is:
62250

Solution:
The osmotic pressure of a non-electrolytic solution can be calculated as follows,

π=CRT

π = osmotic pressure

C = molarity

400 Pa105=2.5 gMo250/1000 L×0.83L-barK.mol×300 K

⇒ 400 × 10-5(1.01325) = 2.5 × 1000Mo × 250 0.083 × 300

⇒ 400 × 10-5 = 10Mo × 0.083 × 300

⇒ Mo = 10 × 0.83 × 300400 × 105

⇒ Mo = 10 × 0.083 × 34 × 105

⇒ Mo=62250 gram/mole

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (31 Jan Shift 1)
At 300oC, FeCl3

Options:
A decomposes into FeCl2 and Cl2 .
B decomposes into Fe and Cl2 .
C sublimes go give liquid FeCl3
D sublimes to give gaseous dimmer FeCl32 .

The correct answer is:
sublimes to give gaseous dimmer FeCl32 .

Solution:
True statement, as at 300oC, FeCl3 sublimes to give gaseous dimmer as follows
2FeCl3â300oCFeCl32g

Topic: Chemistry, d and f Block Elements, JEE Main
At 30 °C, the half life for the decomposition of AB2 is 200 s and is independent of the initial concentration of AB2. The time required for 80% of the AB2 to decompose is (Given: log2=0.30; log3=0.48)

Options:
A 200s
B 323s
C 467s
D 532s

The correct answer is:
467s

Solution:
Since half life is independent of intial concentration of the reactant, it's a first order reaction

k=ln 2t12
kt=2.303 logA0A

ln 2t12t=2.303 logA0A

2.303 log 2200t=2.303 log10020

t=2000.3×log 5=466.67 sec≈467 sec

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2022 (26 Jul Shift 2)
At 40°C, a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from 40°C to 20°C it regains its original length of 0.2 m. The value of M is close to:
(Coefficient of linear expansion and Young’s modulus of brass are 10-5/°C and 1011 N/m2, respectively; g=10 m s-2 )

Options:
A 0.9 kg
B 0.5 kg
C 1.5 kg
D 9 kg

The correct answer is:
9 kg

Solution:
Mg=AYlΔl
Δll=αΔT
Mg=AYαΔT
M=AYαΔTg=π×0.0012×1011×10-5×2010=2π
Closest value of M is 9 kg.

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2019 (12 Apr Shift 1)
At 48°C the vapour pressure of pure C S 2 is 850 torr. A solution of 2.0  g of sulphur in 100 g of C S 2 has a vapour pressure 844.9 torr. Determine the atomicity of sulphur molecule :

Options:
A 1
B 2
C 4
D 8

The correct answer is:
8

Solution:
Δ P = 5.1 torr ; x is atomicity of sulphur.
Δ P P A â = 5.1 850 = X B = 2 3 2 x 2 3 2 x + 1 0 0 7 6
5.1 850 ≅ 2 3 2 x × 7 6 1 0 0
x = 2 × 7 6 3 2 × 1 0 0 × 8 5 0 5.1 = 7.92 ≈ 8

Topic: Chemistry, Solutions, JEE Main
At 100oC and 1 atm. if the density of liquid water is 1.0 gcm-3 and that of water vapour is 0.0006 gcm-3 , then the volume occupied by water molecules in 1 litre of steam at that temperature

Options:
A 6 cm3
B 60 cm3
C 0.6 cm3
D 0.06 cm3

The correct answer is:
0.6 cm3

Solution:
1L=1000 mL=1000 cm3
mass=density×volume
=(0.0006 g cm-3)×(1000 cm3)=0.6g
As 18 g of water =18 cm3
So 0.6 g of water =0.6 cm3
Hence actual volume occupied by molecules =0.6 cm3

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
At 100oC, copper Cu has FCC unit cell structure with cell edge length of x Å . What is the approximate density of Cu (in g cm-3 ) at this temperature?
Atomic Mass of Cu=63.55 u

Options:
A 105x3
B 205x3
C 422x3
D 211x3

The correct answer is:
422x3

Solution:
Crystal structure is FCC, so number of atoms per unit cell N=4
Edge length =x.Å=x 10-8 cm3
Density ZMNAa3=4×63.56.03×1023x×10-83gm/cm3
=4×63.56.0×x3×10-1=422x3gm/cm3=422x3

Topic: Chemistry, Solid State, JEE Main, JEE Main 2019 (09 Jan Shift 2)
At 273 K and one atm, âaâ litre of N2O4 decomposes to NO2 as : N2O4gâ2NO2g. To what extent has the decomposition proceeded when he original volume is 25% less then hat of existing volume?
[Report your answer up to two decimal places.]

Options:

The correct answer is:
33.33

Solution:
âµMolesâvolume a consant P and T
Initial MolesMoles at equilibrium N2O4a(a-x)â2NO2(g)02x
Given original volume =75100 of existing volume at equilibrium
or initial moles = 75 100 Ãmoles at equilibrium
â´a=75100(a+x)
or x=2575a=0Â·33.33a
â´Degree of dissociation, Î±=xa
= 0.3333a a =0.3333âorâ33.33%

Topic: Chemistry, Chemical Equilibrium, JEE Main
At 277 K, degree of dissociation water is 1×10-7% . The value of ionic product of water is

Options:
A 3.0×10-14
B 3.085×10-15
C 1×10-16
D 1×10-14

The correct answer is:
3.085×10-15

Solution:
The molar concentration of water
=1000Ã118=55.5 M
Î± , the degree of dissociation =1Ã10-7100
=1Ã10-9
Kw=CÎ±ÃCÎ± or C2Î±2
Kw=(55.5)2(1Ã10-9)2=3.085Ã10-15

Topic: Chemistry, Ionic Equilibrium, JEE Main
At \( 277 \mathrm{~K} \), degree of dissociation water is \( 1 \times 10^{-7} \% \). The value of ionic product of water is

Options:
A \( 3.0 \times 10^{-14} \)
B \( 3.085 \times 10^{-15} \)
C \( 1 \times 10^{-16} \)
D \( 1 \times 10^{-14} \)

The correct answer is:
\( 3.085 \times 10^{-15} \)

Solution:
The molar concentration of water
=1000Ã118=55.5 M
Î± , the degree of dissociation =1Ã10-7100
=1Ã10-9
Kw=CÎ±ÃCÎ± or C2Î±2
Kw=(55.5)2(1Ã10-9)2=3.085Ã10-15

Topic: Chemistry, Ionic Equilibrium, JEE Main
At 298.2 K the relationship between enthalpy of bond dissociation (in kJ mol-1) for hydrogen EH and its isotope, deuterium ED, is best described by:

Options:
A EH=12ED
B EH=ED
C EH=ED-7.5
D EH=2ED

The correct answer is:
EH=ED-7.5

Solution:
Enthalpy of bond dissociation kJ/mole at 298.2 K

For, hydrogen =435.88

For, Deuterium =443.35

∴ EH≃ED-7.5

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (25 Jul Shift 1)
At 298 K

N2g+3H2g⇌2NH3g,K1=4×105

N2g+O2g⇌2NOg,K2=1.6×1012

H2g+12O2g⇌H2O(g),K3=1.0×10-13

Based on above equilibria, the equilibrium constant of the reaction,

2NH3g+52O2( g)⇌2NOg+3H2Og
is _____ ×10-33 (Nearest integer)

Options:

The correct answer is:
4

Solution:
Given,
N2g3H2g⇌2NH3g,K1=4×105 …1

N2g+O2g⇌2NOg,K2=1.6×1012 …2

H2g+12O2g⇌H2Og,K3=1.0×10-13 …3
Using properties of equilibrium constant:

2+3×3-1

2NH3g+52O2g⇌2NOg+3H2Og

Keq=K2×K33K1=1.6×1012×10-1334×105

=1.64×10-32=4×10-33

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2023 (29 Jan Shift 2)
At 298 K, the enthalpy of fusion of a solid X is 2.8 kJ mol-1 and the enthalpy of vaporisation of the liquid X is 98.2 kJ mol-1. The enthalpy of sublimation of the substance X in kJ mol-1 is ______. (in nearest integer)

Options:

The correct answer is:
101

Solution:
ΔHsub=ΔHfus.+ΔHvap.

=2.8+98.2

=101 kJ/mol

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (25 Jul Shift 1)
At 298 K, the equilibrium constant is 2×1015 for the reaction:

Cus+2Ag+aq⇌Cu2+aq+2Ags

The equilibrium constant for the reaction

12Cu2+aq+Ags⇌12Cus+Ag+aq is x×10-8. The value of x is ______. (Round off the answer to the nearest integer)

Options:

The correct answer is:
2

Solution:
Cus+2Ag+aq⇌Cu2+aq+2Ags; Keq=2×1015

12Cu2+aq+Ags⇌12Cus+Ag+aq

The equilibrium reaction is reversed and multiplied with 12,

Hence, Keq'=1Keq=12×1015=x×10-8

⇒120×1107=x×10-8

⇒120×10-7=x×10-8

1020=x

⇒x=102=5=2.236

≃2.24

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2022 (26 Jul Shift 1)
At 298 K, the standard reduction potentials are 1.51 V for MnO4- | Mn2+, 1.36 V for Cl2|Cl-, 1.07 V for Br2|Br-, 0.54 V for I2|I-. At pH=3, permanganate is expected to oxidize: RTF=0.059

Options:
A Cl- and Br-
B Br- and I-
C I- only
D Cl- , Br- and I-

The correct answer is:
Br- and I-

Solution:
MnO4-+8H++5e-→Mn2++4H2O
EMnO4-Mn2+=Eo-0.0595log⁡Mn2+MnO4-H+8
=1.51-0.0595log⁡110-38
(Assuming MnO4-=Mn2+=1M )
=1.51-0.0595×24=1.51-0.28
=1.23 V
EredMnO4-Mn2+o=1.23 V>EredBr2Br-o>EredI2I-o
i.e. it will oxidise Br- and I-, only.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2015 (11 Apr Online)
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 bar at the same temperature, what would be its concentration?
[Report your answer upto two decimal places.]

Options:

The correct answer is:
0.06

Solution:
π = CRT = W B × R × T M B × V
For both solutions, R, T and V are constants.
For I solution
4.98 bar = 36g × R × T 180 g mol - 1 × V ...(i)
For II solution
1.52 bar = W B × R × T M B × V ...(ii)
On dividing Eq. (ii) by Eq. (i), we get
1.52 bar 4.98 bar = W B × R × T M B × V × 180 × V 36 × R × T
W B M B = 1.52 4.98 × 5 = 0.0610 mol L - 1

Topic: Chemistry, Solutions, JEE Main
At 300 K,

A ⇌  Product    Δ G T o =−200  kJmol −1

B ⇌  Product    Δ G T o =−50  kJmol −1

Thus, the ratio of equilibrium constant at 300 K is

Options:
A 100
B 1000
C 10000
D none of these

The correct answer is:
10000

Solution:
Î G A â = - 2.303 RT log 1 0 K A
Î G B â = - 2.303 RT log 1 0 K B
â´ Î G A â Î G B â = log 1 0 K A K B
4 = log 1 0 K A K B
â´ K A K B = 1 0 4 = 1 000 0

Topic: Chemistry, Chemical Equilibrium, JEE Main
At 300 K, a sample of 3.0 g of gas A occupies the same volume as 0.2 g of hydrogen at 200 K at the same pressure. The molar mass of gas A is g mol-1. (nearest integer) Assume that the behaviour of gases as ideal. (Given: The molar mass of hydrogen H2 gas is 2.0 g mol-1.)

Options:

The correct answer is:
45

Solution:
PV=nRTV=nRTPVH2=VA (Given)

nRTPA=nRTPH2

3×300MA=0.22×200

Hence, the molar mass of A, MA=45g

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2022 (24 Jun Shift 2)
At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume, for complete combustion. After combustion, the gases occupy 345 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is:
(Assume complete combustion of reactant)

Options:
A C4H8
B C4H10
C C3H6
D C3H8

The correct answer is:
C3H8

Solution:
Volume of N2 in air=375×0.8=300 ml
Volume of O2 in air=375×0.2=75 ml
C x H y + x + y 4 O 2 ⟶ xCO 2 g + y 2 H 2 O ℓ 1 5 ml 1 5 x + y 4 0 0 1 5 x -
After combustion, total volume
345=VN2+VCO2
345=300+15x
x=3
Volume of O2 used
15x+y4=75
x+y4=5
y=8
So, hydrocarbon isC3H8.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2016 (03 Apr)
At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O2 for complete combustion, and 40 mL of CO2 is formed. The formula of the hydrocarbon is:

Options:
A C4H8
B C4H10
C C4H6
D C4H7Cl

The correct answer is:
C4H6

Solution:
Let hydrocarbon is CxHy
10x=40 ; x=4 and 10x+y4=55
x+y4=5510
x+y4=5.5
Put the value of x
4+y4=5.5
y4=5.5-4
y4=1.5 then y=6
Hence hydrocarbon is C4H6

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2019 (10 Apr Shift 1)
At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen N2 at 4 bar. The molar mass of the gaseous molecule is

Options:
A 224 g mol-1
B 112 g mol-1
C 56 g mol-1
D 28 g mol-1

The correct answer is:
112 g mol-1

Solution:
d=PMwtRTd1P1M1=d2P2M2
2d22×Mgas=d24×MN2
Mgas=4×MN2
=112 g mol−1.

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2017 (09 Apr Online)
At 300 k the half‐life a sample of a gaseous compound initially at 1 atm is 100 sec. When the pressure is 0.5 atm the half‐life is 50 sec. The order of reaction is

Options:

The correct answer is:
0

Solution:
For a zero order reaciton, t 1/2 ∝initial concentration
⇒ t 1/2 ∝pressure
Since, half life decreases to half with similar decrease in pressure. Hence, the order of reaction is zero.

Topic: Chemistry, Chemical Kinetics, JEE Main
At 300 K, the rms speed of oxygen molecules is α+5α times to that of its average speed in the gas. Then, the value of α will be (use π=227)

Options:
A 32
B 28
C 24
D 27

The correct answer is:
28

Solution:
RMS speed is given by, vrms=3RTM and average speed is given by, vavg=8πRTM.

As per question,

vrms=α+5αvavg

⇒3RTM=α+5α8πRTM

⇒3=α+5α×8227

⇒α=28

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2023 (29 Jan Shift 2)
At 300 K, the vapour pressure of a solution containing 1 mole of n -hexane and 3 moles of n -heptane is 550 mm of Hg. At the same temperature, if one more mole of n -heptane is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. What is the vapour pressure in mmHg of n - heptane in its pure state______?

Options:

The correct answer is:
600

Solution:
Ptotal=P∘hexane. Xhexane+P∘heptane.Xheptane

550=P∘hexane×14+P∘heptane×34 ..(i)

After mixing 1 mole n-heptane

560=P∘hexane×15+P∘heptane×45 .....(ii)

On solving eq. (i) and (ii)

P∘heptane=600 mm of Hg

Topic: Chemistry, Solutions, JEE Main, JEE Main 2020 (04 Sep Shift 1)
At 320 K, a gas A2 is 20 % dissociated to A(g) . The standard Gibbs free energy change at 320 K and 1 atm in J mol–1 is approximately: (R = 8.314 JK–1 mol–1 ;ln2 = 0.693 ;ln3 = 1.098)

Options:
A 1844
B 2068
C 4281
D 4763

The correct answer is:
4763

Solution:
Let assume initial moles is 1
A2→ 2A
t=0 1
t=t 1-0.2 0.2×2
Partial pressure of A2=0.81.2×1
Partial pressure of A=0.41.2×1
kp=0.41.22×10.81.2×1 =16
∆G°= -RTlnk=-8.314×320 ln16
=8.314×320 0.693+1.098
⋍4763 J/mole

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2018 (16 Apr Online)