datavorous/entrance-exam-dataset huggingface dataset:
30 mL of 0.2 N BaCl2 is mixed with 40 mL of 0.3 N Al2SO43 . How many g of BaSO4 are formed?

Options:
A 0.10 g
B 0.60 g
C 0.90 g
D 0.70 g

The correct answer is:
0.70 g

Solution:
BaCl 2 + Al 2 SO 4 3 → BaSO 4 + AlCl 3 Meq. Before Reaction 3 0 × 0 . 2 4 0 × 0 . 3 0 0 = 6 = 1 2 = 0 = 0 Meq. After Reaction 0 6 6 6
6Â Meq.Â ofÂ BaCl2 reacts with 6Â Meq.Â ofÂ Al2SO43 to produce 6Â Meq.Â ofÂ BaSO4 and 6Â Meq.Â ofAlCl3.Â
â´Â Â Â Meq.Â ofÂ BaSO4Â formedÂ =Â 6Â =Wt.EÂ ÃÂ 1000; E=2332
wt.Â ofÂ BaSO4Â formedÂ =6Â ÃE1000=6Â Ã2332Â ÃÂ 1000=Â Â 0.70Â gÂ Â

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
\( 0.30 \mathrm{~g} \) of an organic compound containing \( \mathrm{C}, \mathrm{H} \) and Oxygen on combustion yields \( 0.44 \mathrm{~g} \mathrm{CO}_{2} \)
and \( 0.18 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} \). If one mol of compound weighs \( 60 \), then molecular formula of the compound
is

Options:
A \( \mathrm{CH}_{2} \mathrm{O} \)
B \( \mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O} \)
C \( \mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O} \)
D \( \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2} \)

The correct answer is:
\( \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2} \)

Solution:
Percentage of $C=\frac{12}{44} \times \frac{0.44}{0.30} \times 100=40 \%$
Percentage of $\mathrm{H}=\frac{2}{18} \times \frac{0.18}{0.30} \times 100=6.6 \%$
Percentage of $O=100-(40+6.6)=53.4 \%$
\begin{array}{|l|l|l|l|}
\hline Element & \% & Molar ratio & Simplest ratio \\
\hline \mathrm{C} & 40 & \frac{40}{12}=3.3 & \frac{3.3}{3.3}=1 \\
\hline \mathrm{H} & 6.6 & \frac{6.6}{1}=6.6 & \frac{6.6}{3.3}=2 \\
\hline \mathrm{O} & 53.4 & \frac{53.4}{16}=3.3 & \frac{3.3}{3.3}=1 \\
\hline
\end{array}
hence, empirical formula $=\mathrm{CH}_{2} \mathrm{O}$
We know
$n=\frac{\text { Molecular mass }}{\text { Empirial formula mass }}=\frac{60}{30}=2$ $\Rightarrow$ Molecular formula of compound $=\left(\mathrm{CH}_{2} \mathrm{O}_{2}\right)=\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$ Hence, molar mass of $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}=2 \times 12+1 \times 4+16 \times 2=60 \mathrm{gmol}^{-1}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
30 mL of 0.1 M K I (aq) and 10 mL of 0.2 M AgN O 3 are mixed. The solution is then filtered out. Assuming no change in total volume, depression in freezing point of the resulting solution will be : (Given : Kf for H2O = 1.86 K kg mol-1, assume molarity = molality)

Options:

The correct answer is:
0.28

Solution:
KI(aq) + AgNO 3 (aq) â¶ KNO 3( (aq) + AgI(s) millimoles 3 0 Ã 0 . 1 1 0 Ã 0 . 2 3 2 0 0 mm left 1 0 2 2
â´ [KI] in solution = 1 4 0 and [KNO3] = 2 4 0 &
(Assuming molarity = molality for dilute soluation)
â´ââÎ T f â=âÎ T f ââ(byâKI)ââ+ââÎ T f ââ(byâKN O 3 )
= molality Ã 1.86 Ã i KI + molality Ã 1.86 Ã i KNO 3
=( 1 40 Ã1.86Ã2 )+( 2 40 Ã1.86Ã2 )
= 0.093 + 0.186 = 0.279 = 0.28 Â°C

Topic: Chemistry, Solutions, JEE Main
30 ml of 0.2 M NaOH is added with 50 ml 0.2 M CH3COOH solution. The extra volume of 0.2 M NaOH required to make the pH of the solution 5.00 is 10x .The value of x is. The ionisation constant of CH3COOH=2×10-5 .

Options:

The correct answer is:
3

Solution:
Milli.moles of NaOH=30Ã0.2=6
m.m of CH3COOH=50Ã0.2=10
Now:
pH=-logâ¡(2Ã10-5) +logâ¡64
=4.87
Suppose 'v'mL of NaOH is added then
m.mol of CH3COONa=6+vÃ0.2
m.mol of CH3COOH=4-vÃ0.2
pH= pK a +log Salt Acid
5=-logâ¡(2Ã10-5) +logâ¡6+0.2v4-0.2v
0.3010=logâ¡6+0.2v4-0.2v
So 6+0.2v4-0.2v=2âv=3.33 mL=103mL
So x=3

Topic: Chemistry, Ionic Equilibrium, JEE Main
3.0 molal aqueous solution of an electrolyte A2B3 is 50 % ionised. The boiling point of the solution at 1 atm is: KbH2O=0.52 Kkgmol-1

Options:
A 274.76 K
B 377.68 K
C 374.68 K
D 104.68 K

The correct answer is:
377.68 K

Solution:
A2B3(aq)→2A3+(aq)+3B2-(aq)
n=5
ΔTb=i.Kb.m
=[1+(n-1)α]Kb.m.
=[1+4α]Kb.m.
=[1+4(0.5)]0.52×3
=(1+2)(0.52)×3
=4.68
Ab=373+4.68=377.68K

Topic: Chemistry, Solutions, JEE Main
32 g of O2 is contained in a cubical container of side 1 m and maintained at a temperature of 127 Â°C. The isothermal bulk modulus of elasticity of the gas is (universal gas constant = R)

Options:
A 127 R

B 400 R

C 200 R

D 560 R

The correct answer is:
400 R

Solution:
The bulk modulus of elasticity at a constant temperature
PV=nRT
dPP=dVV
B= -dPdV/V
B=P
B=nRTV= R40013=400R

Topic: Physics, Mechanical Properties of Solids, JEE Main
32 g of O2 is contained in a cubical container of side 1 m and maintained at a temperature of 127oC. The isothermal bulk modulus of elasticity of the gas in terms of universal gas constant R is

Options:
A 127 R
B 400 R
C 200 R
D 560 R

The correct answer is:
400 R

Solution:
Bulk modulus of elasticity at constant temperature
B=-dPdV/V=-VdPdVT=+P=nRTV
=1×R×40013=400R

Topic: Physics, Mechanical Properties of Solids, NEET
40 g of glucose (Molar mass =180) is mixed with 200 mL of water. The freezing point of solution is __K. (Nearest integer)
[Given : Kf=1.86 K kg mol-1; Density of water =1.00 g cm-3; Freezing point of water =273.15 K

Options:

The correct answer is:
271

Solution:
Molality =40180mol0.2Kg=109 molal

⇒ΔTf=Tf-Tf'=1.86×109

⇒Tf'=273.15-1.86×109

=271.08 K

≈271 K (nearest-integer)

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (27 Aug Shift 2)
0.40 g of helium in a bulb at a temperature of $T \mathrm{~K}$ had a pressure of $p$ atm. When the bulb was immersed in water bath at temperature 50 K more than the first one, 0.08 g of gas had to be removed to restore the original pressure. $T$ is

Options:
A 500 K
B 400 K
C 600 K
D 200 K

The correct answer is:
200 K

Solution:
As pressure and volume remains constant,
$\begin{aligned} n_1 T_1 & =n_2 T_2 \\ n_1 & =\frac{0.4}{4}=0.1, T_1=T \mathrm{~K} \\ n_2 & =\frac{0.40-0.08}{4}=0.08, T_2=(T+50) \mathrm{K}\end{aligned}$
On putting the values,
$\begin{array}{ll} & 0.1 \times T=0.08 \times(T+50) \\ \text { or } & 0.1 \mathrm{~T}=0.08 T=4 \text { or } 0.1 \mathrm{~T}-0.08 \mathrm{~T}=4 \\ \text { or } & 0.02 T=4 \text { or } T=200 \mathrm{~K}\end{array}$

Topic: Chemistry, States of Matter, JIPMER, JIPMER 2015
40% of HI undergoes decomposition to H2 and I2 at 300 K.ΔGΘ for this decompostion reaction at one atmopsphere pressure is____Jmol-1-[nearest integer]

Use R=8.31JK-1 mol-1;log2=0.3010,ln10=2.3,log3=0.477

Options:

The correct answer is:
2735

Solution:
HIg⇌12H2g+12I2g

Initial moles	1	0	0

moles at equilibrium	1-0.4	0.2	0.2

KP=0.212×0.2120.6

KP=0.20.6=13

ΔG°=-2.3RTlogKP

=-2.3×8.31×300log13

=2375Jmol-1

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2022 (26 Jun Shift 2)
40% (w/v) of NaCl solution (specific gravity = 1.12) is equivalent to

Options:
A 3.57×105 ppm
B 3.57×106 ppm
C 1×106 ppm
D 4×105 ppm

The correct answer is:
3.57×105 ppm

Solution:
Volume of solution = 100 mL
Mass of solution = Volume × density
Mass of solution =100×1.12=112 g
Concentration in ppm =WsoluteWsolution×106
ppm in NaCl solution =40112×106
=3.57×105 ppm.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
42.12% (w/v) solution of NaCl causes precipitation of a certain sol in 10 hours. The coagulating value of NaCl for the sol is
[Given : Molar mass : Na=23.0 g mol-1; Cl=35.5 g mol-1]

Options:
A 36 mmolL-1
B 36 mol L-1
C 1440 mol L-1
D 1440 m molL-1

The correct answer is:
1440 m molL-1

Solution:
The minimum concentration of an electrolyte in millimoles per litre required to cause precipitation of a sol in two hours is called coagulating value.

42.12 % means 100 ml solution contain 42.12 g of salt.moles of salt = 42.1258.5 = 0.72 moles = 720 milimoles

Since 100 ml required 720 milimoles

therefor 1 L will required 7200 milimoles.

It will coagulate in 10 hours so for 2 hours = 5×7200 = 14400 mili mol L-1

Note: NTA has either made mistake in question data or in options.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2022 (29 Jun Shift 2)
\( 0.44 \mathrm{~g} \) of a monohydric alcohol when added to methylmagnesium iodide in ether liberates at
S.T.P., \( 112 \mathrm{~cm}^{3} \) of methane. With PCC the same alcohol forms a carbonyl compound that
answers silver mirror test. The monohydric alcohol is

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

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

Solution:
At STP $112 \mathrm{~cm}^{3}$ of methane liberated, $22400 \mathrm{~cm}^{3}=1 \mathrm{~mol}$ at ST
$1 \mathrm{~cm}^{3}=\frac{1}{22400} \mathrm{~mol}$
$112 \mathrm{~cm}^{3}=\frac{1}{22400} \times 112 \mathrm{~mol}$
$=0.005$ mol of methane liberated
1 mol of monohydric alcohol react with 1 mol of methyl magnesium iodide to produces $1 \mathrm{~mol}$ of methane, according to
the reaction,
Monohydric alcohol+ $\mathrm{CH}_{3} \mathrm{M} \mathrm{gl}-\overline{\text { Either }}$
$\Rightarrow 0.005$ mol of monohydric alcohol produces $0.005$ mol of methane
Now, molar mass of monohydric alcohol
$=\frac{\text { Given mass of monohydric alcohol }}{\mathrm{Number} \text { of moles of monohydric alcohol }}$
$=\frac{0.44}{0.005}=88 \mathrm{~g}$
Monohydric alcohol from the given option having $88 \mathrm{~g}$ molar mass is $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}$
$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{M} \mathrm{gI}-\mathrm{Either}$
When react with $\mathrm{PCC}$, aldehyde is formed which give positive silver mirror test.
( $\mathrm{CH}_{3}$ ) $\left._{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}-\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OM} \mathrm{gI}+\mathrm{CH}_{4}$
( $\mathrm{CH}_{3}$ ) $_{3} \mathrm{C}-\mathrm{CHO}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
0.45 gm of an organic compound gave on combustion 0.792 gm of CO2 and 0.324 gm of water. 0.24 gm of the same substance was Kjeldahlised and the ammonia liberated was absorbed in 50.0 ml of M/8 H2SO4. The excess acid required 77.0 ml of N/10 NaOH for complete neutralisation. Calculate the empirical formula of the compound.

Options:
A H4C2N8O
B C2H4N8O
C C4H8N2O
D C2N2H4O

The correct answer is:
C4H8N2O

Solution:
Calculation of the percentage composition :
i) Percentage of carbon
= 1 2 4 4 × Mass of CO 2 Produced Mass of substance taken × 1 0 0
= 1 2 4 4 × 0 . 7 9 2 0 . 4 5 × 1 0 0 = 4 8 . 0
ii) Percentage of hydrogen
= 2 1 8 × Mass of H 2 O produced Mass of substance taken × 1 0 0
= 2 1 8 × 0 . 3 2 4 0 . 4 5 × 1 0 0 = 8 . 0
iii) Percentage of nitrogen
mEq. of H2SO4 = 5 0 × M 8 × 2 = 1 2 . 5
mEq. of NaOH = 7 2 × 1 1 0 × 1 = 7 . 7
Excess of H2SO4 used to neutralise ammonia
= 12.5 - 7.7 = 4.8
Percentage of N2 = 1 . 4 × 4 . 8 0 . 2 4 = 2 8 . 0 %
iv) Percentage of oxygen = 100 - (percentage of C + percentage of H + percentage of N) = 100 - (48.0 + 8.0 + 28.0) = 16.0.
b. Calculation of empirical formula :

Element	Percentage	Atomic mass number of atoms	Relative number of atoms	Simlpest atomic ratio	Simplest whole number atomic ratio

Carbon	48.8	12	4 8 . 0 1 2 = 4	4 1 = 4	4

Hydrogen	8.0	1	8 . 0 1 = 8	8 . 0 1 = 8	8

Nitrogen	28.0	14	2 8 . 0 1 4 = 2	2 1 = 2	2

Oxygen	16.0	16	1 6 . 0 1 6 = 1	1 1 = 1	1

Hence, the empirical formula of the compound is C4H8N2O.

Topic: Chemistry, Practical Chemistry, JEE Main
46. If $M(A ; Z), M_p$ and $M_n$ denote the masses of the nucleus ${ }_Z^A X$, proton and neutron respectively in units of $\mathrm{u}\left(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{C}^2\right)$ and $\mathrm{BE}$ represents its bonding energy in $\mathrm{MeV}$, then

Options:
A $M(A, Z)=Z M_p+(A-Z) M_n+\mathrm{BE} / C^2$
B $M(A, Z)=Z M_p+(A-Z) M_n-\mathrm{BE} / \mathrm{C}^2$
C $$
\begin{aligned}
& M(A, Z)=Z M_p+(A-Z) M_n+B E \\
\end{aligned}
$$
D $$
\begin{aligned}
& M(A, Z)=Z M_p+(A-Z) M_n-B E
\end{aligned}
$$

The correct answer is:
$M(A, Z)=Z M_p+(A-Z) M_n-\mathrm{BE} / \mathrm{C}^2$

Solution:
$$
\mathrm{BE}=\left[\mathrm{ZM}_{\mathrm{p}}+(\mathrm{A}-\mathrm{Z}) \mathrm{M}_{\mathrm{n}}-\mathrm{M}(\mathrm{A}, \mathrm{Z})\right] \mathrm{C}^2
$$

Topic: Physics, Nuclear Physics, NEET, NEET 2008 (Mains)
$50 \mathrm{~cm}^{3}$ of $0.2 \mathrm{~N} \mathrm{HCl}$ is titrated against $0.1 \mathrm{~N}$ $\mathrm{NaOH}$ solution. The titration is discontinued after adding $50 \mathrm{~cm}^{3}$ of $\mathrm{NaOH}$. The remaining titration is completed by adding $0.5 \mathrm{~N} \mathrm{KOH}$. The volume of $\mathrm{KOH}$ required for completing the titration is

Options:
A $12 \mathrm{~cm}^{3}$
B $10 \mathrm{~cm}^{3}$
C $25 \mathrm{~cm}^{3}$
D $10.5 \mathrm{~cm}^{3}$

The correct answer is:
$10 \mathrm{~cm}^{3}$

Solution:
When $0.1 \mathrm{~N} \mathrm{NaOH}$ is used,
$$
\begin{gathered}
\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2} \\
\text { (For HCl) } \\
\text { (For NaOH) } \\
0.2 \mathrm{~N} \times \mathrm{V}_{1}=50 \times 0.1 \mathrm{~N} \\
\mathrm{~V}_{1}=\frac{50 \times 0.1}{0.2}=25 \mathrm{~cm}^{3}
\end{gathered}
$$
When $0.5 \mathrm{~N} \mathrm{KOH}$ is used,
$$
\begin{aligned}
\mathrm{N}_{1} \mathrm{~V}_{1}=& \mathrm{N}_{3} \mathrm{~V}_{3} \\
\text { (For remaining HCl) } &(\text { For KOH}) \\
0.2 \mathrm{~N} \times 25 &=0.5 \mathrm{~N} \times \mathrm{V}_{3} \\
\mathrm{~V}_{3} &=\frac{0.2 \times 25}{0.5} \\
=& 10 \mathrm{~cm}^{3}
\end{aligned}
$$

Topic: Chemistry, Redox Reactions, JEE Main
$50 \mathrm{~cm}^{3}$ of $0.2 \mathrm{~N} \mathrm{HCl}$ is titrated against $0.1 \mathrm{~N}$ $\mathrm{NaOH}$ solution. The titration is discontinued after adding $50 \mathrm{~cm}^{3}$ of $\mathrm{NaOH}$. The remaining titration is completed by adding $0.5 \mathrm{~N} \mathrm{KOH}$. The volume of $\mathrm{KOH}$ required for completing the titration is

Options:
A $12 \mathrm{~cm}^{3}$
B $10 \mathrm{~cm}^{3}$
C $25 \mathrm{~cm}^{3}$
D $10.5 \mathrm{~cm}^{3}$

The correct answer is:
$10 \mathrm{~cm}^{3}$

Solution:
When $0.1 \mathrm{~N} \mathrm{NaOH}$ is used,
$$
\begin{gathered}
\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2} \\
\text { (For HCl) } \\
\text { (For NaOH) } \\
0.2 \mathrm{~N} \times \mathrm{V}_{1}=50 \times 0.1 \mathrm{~N} \\
\mathrm{~V}_{1}=\frac{50 \times 0.1}{0.2}=25 \mathrm{~cm}^{3}
\end{gathered}
$$
When $0.5 \mathrm{~N} \mathrm{KOH}$ is used,
$$
\begin{aligned}
\mathrm{N}_{1} \mathrm{~V}_{1}=& \mathrm{N}_{3} \mathrm{~V}_{3} \\
\text { (For remaining HCl) } &(\text { For KOH}) \\
0.2 \mathrm{~N} \times 25 &=0.5 \mathrm{~N} \times \mathrm{V}_{3} \\
\mathrm{~V}_{3} &=\frac{0.2 \times 25}{0.5} \\
=& 10 \mathrm{~cm}^{3}
\end{aligned}
$$

Topic: Chemistry, Redox Reactions, KCET, KCET 2010
$50 \mathrm{~cm}^{3}$ of $0.2 \mathrm{~N} \mathrm{HCl}$ is titrated against $0.1 \mathrm{~N}$ $\mathrm{NaOH}$ solution. The titration was discontinued after adding $50 \mathrm{~cm}^{3}$ of $\mathrm{NaOH}$. The remaining titration is completed by adding $0.5 \mathrm{~N} \mathrm{KOH}$. The volume of $\mathrm{KOH}$ required for completing the titration is

Options:
A $12 \mathrm{~cm}^{3}$
B $10 \mathrm{~cm}^{3}$
C $21.0 \mathrm{~cm}^{3}$
D $16.2 \mathrm{~cm}^{3}$

The correct answer is:
$10 \mathrm{~cm}^{3}$

Solution:
No. of equivalent of $\mathrm{HCl}$ remaining after adding
$$
\begin{aligned}
50 \mathrm{~cm}^{3} \text { of } 0.1 \mathrm{~N} \mathrm{NaOH} &=\frac{0.2 \times 50-0.1 \times 50}{100} \\
&=\frac{0.5}{100}
\end{aligned}
$$
$\therefore$ Volume of $0.5 \mathrm{~N} \mathrm{KOH}$ required $\frac{0.5}{100} \mathrm{eq}$
$$
\begin{aligned}
& \equiv \frac{V \times 0.5}{1000} \\
V &=\frac{0.5}{100} \times \frac{1000}{0.5} \\
&=10 \mathrm{~cm}^{3}
\end{aligned}
$$

Topic: Chemistry, Redox Reactions, JEE Main
$50 \mathrm{~cm}^{3}$ of $0.2 \mathrm{~N} \mathrm{HCl}$ is titrated against $0.1 \mathrm{~N}$ $\mathrm{NaOH}$ solution. The titration was discontinued after adding $50 \mathrm{~cm}^{3}$ of $\mathrm{NaOH}$. The remaining titration is completed by adding $0.5 \mathrm{~N} \mathrm{KOH}$. The volume of $\mathrm{KOH}$ required for completing the titration is

Options:
A $12 \mathrm{~cm}^{3}$
B $10 \mathrm{~cm}^{3}$
C $21.0 \mathrm{~cm}^{3}$
D $16.2 \mathrm{~cm}^{3}$

The correct answer is:
$10 \mathrm{~cm}^{3}$

Solution:
No. of equivalent of $\mathrm{HCl}$ remaining after adding
$$
\begin{aligned}
50 \mathrm{~cm}^{3} \text { of } 0.1 \mathrm{~N} \mathrm{NaOH} &=\frac{0.2 \times 50-0.1 \times 50}{100} \\
&=\frac{0.5}{100}
\end{aligned}
$$
$\therefore$ Volume of $0.5 \mathrm{~N} \mathrm{KOH}$ required $\frac{0.5}{100} \mathrm{eq}$
$$
\begin{aligned}
& \equiv \frac{V \times 0.5}{1000} \\
V &=\frac{0.5}{100} \times \frac{1000}{0.5} \\
&=10 \mathrm{~cm}^{3}
\end{aligned}
$$

Topic: Chemistry, Redox Reactions, KCET, KCET 2011
50 ml of hydrogen diffuses out through a small hole from a vessel in 20 minutes. Calculate the time (in minutes) needed for 40 ml of oxygen to diffuse out

Options:

The correct answer is:
64

Solution:
If 40 ml O2 will diffuse in t min. then. rO2=40t
rH2=5020
rO2=rH2âMH2MO2=5020232=5020.14
40t=5080â´t=40Ã8050=64 min.

Topic: Chemistry, States of Matter, JEE Main
50 mL of 0.2M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solution is 4.75, the pH of the mixture will be:

Options:
A 8.25
B 9.25
C 3.75
D 4.75

The correct answer is:
9.25

Solution:
NH3          +                      HCl                     ⇌      NH4Clmoles at ​​​​​​​ t=0         50×.2×10−3            25×.2×10−3           −moles at ​​​t=t         25×.2×10−3                       0                           25×.2×10−3
∴ salt & base present in soluton, so it is basic Buffer solution.
pOH=pkb NH3 +log⁡saltbase=4.75∴ pOH=Pkb=4.75

pH=14-pOH=14-4.75=9.25.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2017 (09 Apr Online)
50tan3tan-112+2cos-115+42tan12tan-122 is equal to ______.

Options:

The correct answer is:
29

Solution:
Let S=50tan3tan-112+2cos-115+42tan12tan-122

=50tantan-112+2tan-112+tan-12+42tan12tan-122

Let 12tan-122=α⇒tan2α=22

Now 2tanα1-tan2α=22

Solving we get, tanα=12

SoS=50tantan-112+2·π2+42×12

=50tantan-112+4

=25+4=29

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2022 (29 Jun Shift 1)
$0.53 \mathrm{gm}$ of $\mathrm{Na}_2 \mathrm{CO}_3$ has been dissolved in $100 \mathrm{~m} /$ of a sodium carbonate solution. The normality of the solution will be

Options:
A $\frac{N}{5}$
B $\frac{N}{2}$
C $\frac{N}{10}$
D $\mathrm{N}$

The correct answer is:
$\frac{N}{10}$

Solution:
$N=\frac{0.53 \times 1000}{53 \times 100} \Rightarrow N=\frac{1}{10}$
So normality of the solution will be $\frac{N}{10}$

Topic: Chemistry, Practical Chemistry, JEE Main
\( 5.5 \mathrm{~g} \) of a mixture of \( \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} \) and \( \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 9 \mathrm{H}_{2} \mathrm{O} \) requires \( 5.4 \mathrm{~mL} \) of \( 0.1 \mathrm{~N} \mathrm{KMnO}_{4} \) solution for complete oxidation. Calculate the number of moles of hydrated ferric sulphate in the mixture.

Options:

The correct answer is:
0.0095

Solution:
FeSO4·7H2O will oxidise as Fe is present in +2 oxidation state.

In Fe2SO43·9H2O, Fe is already present in +3 oxidation state, so it will not oxidise.

Hence, milliequivalents of FeSO4·7H2O= milliequivalents of KMnO4

N1V1 FeSO4·7H2O=N2V2 KMnO4x278×1000=0.1×5.4x=0.1501 g

Amount of hydrated ferric sulphate =5.5 g−0.1501 g=5.34 g

Number of moles = Mass / Molar mass =5.34/562=0.0095 mol

Topic: Chemistry, Redox Reactions, JEE Main
\( 5.5 \mathrm{mg} \) of nitrogen gas dissolves in \( 180 \mathrm{~g} \) of water at \( 273 \mathrm{~K} \) and one atm pressure due to
nitrogen gas. The mole fraction of nitrogen in \( 180 \mathrm{~g} \) of water at \( 5 \mathrm{~atm} \) nitrogen pressure is
approximately

Options:
A \( 1 \times 10^{-5} \)
B \( 1 \times 10^{-4} \)
C \( 1 \times 10^{-6} \)
D \( 1 \times 10^{-3} \)

The correct answer is:
\( 1 \times 10^{-4} \)

Solution:
No. of moles of \( N_{2}=\frac{5.5 \times 10^{-3}}{28} \) moles
\[
\begin{array}{l}
=2.36 \times 10^{-4} \text { moles } \\
p_{N_{2}}=k_{H} \times \frac{n_{N_{2}}}{n_{\mathrm{H}_{2} \mathrm{O}}}\left(x_{N_{2}}=\frac{n_{N_{2}}}{n_{\mathrm{H}_{2} \mathrm{O}}} ; n_{\mathrm{H}_{2} \mathrm{O}} \gg n_{N_{2}}\right) \\
1=k_{H} \times \frac{2.36 \times 10^{-4}}{10} \\
k_{H}=\frac{10 \times 10^{4}}{2.36}=4.2 \times 10^{4}
\end{array}
\]
Now, at \( 5 \mathrm{~atm} \) pressure of nitrogen gas, the mole fraction of nitrogen gas is,
\[
\begin{array}{l}
x_{N_{2}}=\frac{p_{N_{2}}}{k_{H}} \\
\Rightarrow x_{N_{2}}=\frac{5}{4.2 \times 10^{4}} \\
\Rightarrow x_{N_{2}}=1.12 \times 10^{-4} \approx 1 \times 10^{-4}
\end{array}
\]

Topic: Chemistry, Amines, JEE Main
0.56 g of a lime stone was treated with oxalic acid to give CaC2O4. The precipitate decolourized 45 mL of 0.2 N  KMnO 4 in acid medium. Calculate % of CaO in the stone.

Options:

The correct answer is:
45

Solution:
The given reaction is
CaCO 3 limeâstone â H 2 C 2 O 4 CaC 2 O 4 â KMnO 4 decolourizes
The redox changes are :
C3+2â2C4++2e
Mn7++5eâMn2+
Meq.âofâCaOâ=âMeq.âofâ CaCO 3 â=âMeq.âofâ H 2 C 2 O 4
=âMeq.âofâ CaC 2 O 4 â=âMeq.âofâ KMnO 4
Hence Meq. of CaO = Meq. of  KMnO 4
w/(56/2)Ã1000=45Ã0.2
So wt.âofâCaOâ=â0.252âg
Here % of CaO in lime stone = 0.2520.256Ã100=45%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$5.6 \mathrm{~L}$ of helium gas at STP is adiabatically compressed to $0.7 \mathrm{~L}$. Taking the initial temperature to be $T_1$, the work done in the process is

Options:
A
$\frac{9}{8} R T_1$

B
$\frac{3}{2} R T_1$

C
$\frac{15}{8} R T_1$

D
$\frac{9}{2} R T_1$

The correct answer is:
$\frac{9}{8} R T_1$

Solution:
At STP, 22.4 L of any gas is 1 mole.
$$
\therefore \quad 5.6 \mathrm{~L}=\frac{5.6}{22.4}=\frac{1}{4} \text { moles }=n
$$
In adiabatic process,
$$
\begin{array}{rlrl}
& T V^{\gamma-1} & =\text { constant } \\
& \therefore & T_2 V_2^{\gamma-1} & =T_1 V_1^{\gamma-1} \\
& \text { or } & T_2 & =T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}
\end{array}
$$
$$
\begin{aligned}
& \gamma=\frac{C_p}{C_V}=\frac{5}{3} \text { for monoatomic He gas. } \\
& \therefore \quad T_2=T_1\left(\frac{5.6}{0.7}\right)^{\frac{5}{3}-1}=4 T_1 \\
& \because \text { Further in adiabatic process, } \\
& Q=0 \\
& \therefore \quad W+\Delta U=0 \\
& \text { or } \quad W=-\Delta U \\
& =-n C_V \Delta T \\
& =-n\left(\frac{R}{\gamma-1}\right)\left(T_2-T_1\right) \\
& =-\frac{1}{4}\left(\frac{R}{\frac{5}{3}-1}\right)\left(4 T_1-T_1\right) \\
& =-\frac{9}{8} R T_1 \\
&
\end{aligned}
$$
$\therefore$ Correct option is (a).
Analysis of Question
(i) From calculation point of view question is moderately tough.
(ii) Volume of gas is decreasing. Therefore, work done by the gas should be negative.

Topic: Physics, Thermodynamics, JEE Main
\( 5.6 \mathrm{~L} \) of helium gas at STP is adiabatically compressed to \( 0.7 \mathrm{~L} \). Taking the initial temperature to be \( T_{1} \) the work done in the process is

Options:
A \( -\frac{9}{8} R T_{1} \)
B \( \frac{3}{2} R T_{1} \)
C \( \frac{15}{8} R T_{1} \)
D \( \frac{9}{2} R T_{1} \)

The correct answer is:
\( -\frac{9}{8} R T_{1} \)

Solution:
At STP, 22.4 L of any gas is equal to 1 mol

∴5.6 L=5.622.4=14 mol=n.

In adiabatic process, TVγ-1= constant

∴ T2V2γ-1=T1V1γ-1 or T2=T1V1V2γ-1

For monoatomic (Helium) gas, γ=CpCV=53

∴T2=T15.60.753-1=4T1

Further, in adiabatic process,

Q=0⇒W+ΔU=0

or W=-ΔU=-nCVΔT

=-nRγ-1T2-T1

=-14R53-14T1-T1=-98RT1

Topic: Physics, Thermodynamics, JEE Main
60 g of gaseous C2H6 are mixed with 28 g of carbon monoxide. The pressure of the resulting gaseous mixture is 3 atm. The partial pressure in atm. of C2H6 in the mixture is

Options:

The correct answer is:
2.00

Solution:
Moles of C2H6=60g32g/mol=2mol
moles of CO=28g28g/mol=1mol
Mole fraction C2H6=22+1=23
Partial pressure of C2H6=PtotalÃ Mole fraction of C2H6
=3atmÃ23=2atm .

Topic: Chemistry, States of Matter, JEE Main