datavorous/entrance-exam-dataset huggingface dataset:
$1.25 \mathrm{~g}$ of a sample of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ and $\mathrm{Na}_{2} \mathrm{SO}_{4}$ is

dissolved in $250 \mathrm{ml}$ solution. $25 \mathrm{ml}$ of this solution neutralises $20 \mathrm{ml}$ of $0.1 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}$. The $\%$ of $\mathrm{Na}_{2} \mathrm{CO}_{3}$

in this sample is

Options:
A $84.8 \%$
B $8.48 \%$
C $15.2 \%$
D $42.4 \%$

The correct answer is:
$84.8 \%$

Solution:
Let the amount of $\mathrm{Na}_{2} \mathrm{CO}_{3}$ present in the mixture be $\mathrm{xg} . \mathrm{Na}_{2} \mathrm{SO}_{4}$ will not react with $\mathrm{H}_{2} \mathrm{SO}_{4} .$ Then

$\frac{x}{53}=\frac{20 \times 0.1 \times 10}{1000} \therefore x=1.06 g$

$\therefore$ Percentage of $\mathrm{Na}_{2} \mathrm{CO}_{3}=$

$\frac{1.06 \times 100}{1.25}=84.8 \%$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
143 calories of heat energy are supplied to a gas. Its internal energy rises by 500 joules. If the pressure remains constant and is equal to 105 N m-2, the change in the volume of the gas is

Options:
A 1000 litres
B 100 litres
C 10 litres
D 1 litre

The correct answer is:
1 litre

Solution:
The energy supplied =143 calories =143Ã4.2 joule=600.6 J du= rise in internal energy =500 J dw = work done = 600.6-500 J
dw=Pdv=100.6 J 105dv=100.6 J
dv=100.6105=10-3 m3=1 litre

Topic: Physics, Thermodynamics, NEET
1.44 g of Titanium (Ti) reacted with an excess of O2 and produced x g of a non-stoichiometric compound Ti1.44O1. The value of x in g is :

Options:
A 1.44
B 2
C 1.77
D None of these

The correct answer is:
1.77

Solution:
Ti + O2 â¶ Ti1.44 O
(1.44 g) (x) g
Apply POAC to Ti
( 1.44g 48 )Ã1={ xg ( 1.44Ã48 )+16 }1.44
x = 4 8 Ã 1 . 4 4 + 1 6 4 8
= 1 . 4 4 + 1 6 4 8
= 1.44 + 0.33
= 1.77 g.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$150 \mathrm{~g}$ of ice is mixed with $100 \mathrm{~g}$ of water at temperature $80^{\circ} \mathrm{C}$. The latent heat of ice is $80 \mathrm{cal} / \mathrm{g}$ and the specific heat of water is $1 \mathrm{cal} / \mathrm{g}-{ }^{\circ} \mathrm{C}$. Assuming no heat loss to the environment, the amount of ice which does not melt is -

Options:
A $100 \mathrm{~g}$
B $0 \mathrm{~g}$
C $150 \mathrm{~g}$
D $50 \mathrm{~g}$

The correct answer is:
$50 \mathrm{~g}$

Solution:
Heat loss by water = heat gain by ice.
$100 \times 1 \times 80=\mathrm{m} \times 80$
$\mathrm{m}=100 \mathrm{gm}$ ice melt
$\therefore$ Remaining ice $=50 \mathrm{~g}$

Topic: Physics, Thermal Properties of Matter, KVPY, KVPY 2012 (SA)
150 g of water at 0 Â°C is contained in a thermally insulated container. Now the air from the vessel is pumped out adiabatically. A fraction of water gets converted into ice and the remaining evaporates at 0 Â°C itself. The mass of evaporated water will be closest to [Take the latent heat of vaporization of water, Lv=2.10Ã106 J kg-1 and latent heat of fusion of water, Lf=3.36Ã105 J kg-1]

Options:
A 20 g
B 130 g
C 35 g
D 150 g

The correct answer is:
20 g

Solution:
Let us assume the mass of the water evaporated is m, then

mLv=150-mLf

m×2.1×106=150-m×3.36×105

On solving we get

m≈20 g

Topic: Physics, Thermal Properties of Matter, JEE Main
164 . The solution of the differential equation $\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0$ is

Options:
A $\tan y \cdot \tan x=C$
B $\frac{\tan y}{\tan x}=C$
C $\frac{\tan ^{2} x}{\tan y}=C$
D None of these

The correct answer is:
$\tan y \cdot \tan x=C$

Solution:
Given, $\frac{\sec ^{2} x}{\tan x} d x=-\frac{\sec ^{2} y}{\tan y} d y$
$$
\Rightarrow \int \frac{\sec ^{2} x}{\tan x} d x=-\int \frac{\sec ^{2} y}{\tan y} d y
$$
Put $\quad \tan x=u \Rightarrow \sec ^{2} x d x=d u$
and $\quad \tan y=v \Rightarrow \sec ^{2} y d y=d v$
$\therefore \quad \int \frac{d u}{u}=-\int \frac{d v}{v}$
$\Rightarrow \log u=-\log v+\log C \Rightarrow w v=C \Rightarrow \tan x \cdot \tan y=C$

Topic: Mathematics, Differential Equations, JEE Main
\(18.4 \mathrm{~g} \mathrm{~N}_2 \mathrm{O}_4\) was placed in \(1 \mathrm{~L}\) vessel at \(400 \mathrm{~K}\) and allowed to attain the following equilibrium \(\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)\).
If the total pressure at equilibrium was 10.64 bar, approximate \(K_p\) is \((R=0.083 \mathrm{~L}\) bar \(\mathrm{K}^{-1} \mathrm{~mol}^{-1}\) ) (Assume \(\mathrm{N}_2 \mathrm{O}_4, \mathrm{NO}_2\) as ideal gases)

Options:
A 57.20
B 24.24
C 14.30
D 6.64

The correct answer is:
24.24

Solution:
From ideal gas equation,
\(\begin{aligned}
p V & =n R T \\
p_{\mathrm{N}_2 \mathrm{O}_4} \times \mathrm{l} & =\frac{18.4}{92} \times 0.083 \times 400 \\
p_{\mathrm{N}_2 \mathrm{O}_4} & =6.64 \text { bar }
\end{aligned}\)
For the equilibrium reaction,
\(\begin{aligned}
p_T & =p-p_i+2 p_i \\
p_T & =p+p_i \Rightarrow p_i=p_T-p \\
p_i & =10.64-6.64 \\
& =4.00 \mathrm{bar}
\end{aligned}\)
At equilibrium,
\(\begin{aligned}
\therefore \quad p_{\mathrm{N}_2 \mathrm{O}_4} & =p-p_i=6.64-4 \\
& =2.64 \mathrm{bar} \\
p_{\mathrm{NO}_2} & =2 p_i=2 \times 4 \\
& =8 \mathrm{bar} \\
\therefore \quad K_p & =\frac{\left[p_{\mathrm{NO}_2}\right]^2}{\left[p_{\mathrm{N}_2 \mathrm{O}_4}\right]} \\
& =\frac{[8]^2}{2.64}=24.29
\end{aligned}\)

Topic: Chemistry, Chemical Equilibrium, JEE Main
200 g water is heated from 40 oC to 60 oC . Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water =4184 J kg-1 K-1):

Options:
A 167.4Â kJ
B 8.4Â kJ
C 4.2Â kJ
D 16.7 kJ

The correct answer is:
16.7 kJ

Solution:
Assuming isochoric process
Q=∆U
∆U=0.2×4184×20
=16.7 kJ

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2016 (09 Apr Online)
$200 \mathrm{mg}$ of ${ }^{14} \mathrm{C}$ will become $25 \mathrm{mg}$ in how many years if the half-life is 5760 years?

Options:
A 5760 years
B 17280 years
C 23040 years
D 11520 years

The correct answer is:
17280 years

Solution:
Half life for radio active ${ }^{19} \mathrm{C}$ is 5760 years.
It is a first order reaction.
$\begin{aligned} & t \frac{1}{2}=\frac{0.693}{k} \\ & \begin{aligned} & k=\frac{0.693}{5760} \\ &=1.203 \times 10^{-4} \text { years }^{-1} \\ & k=\frac{2.303}{t} \log \frac{a}{a-x}\end{aligned} \\ & \begin{aligned} & a=\text { Initial concentration }=200 \mathrm{mg} \\ & a-x=\text { Concentration at time } t=25 \mathrm{mg}\end{aligned}\end{aligned}$
$\begin{aligned} t & =\frac{2.303}{k} \log \frac{200}{25} \\ & =\frac{2.303}{1.203 \times 10^{-4}} \log 8 \\ & =1.914 \times 10^4 \log (2)^3 \\ & =1.914 \times 3 \times 10^4 \times 0.301 \\ & =1.7280 \times 10^4 \\ & =17280 \text { years. }\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main
200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is -57.1 kJ. The increase in temperature in C∘ of the system on mixing is x×10-2. The value of x is (Nearest integer)

[Given: Specific heat of water =4.18 J g-1 K-1

Density of water=1.00 g cm-3]

(Assume no volume change on mixing)

Options:

The correct answer is:
82

Solution:
milimole HCl400+NaOH300⟶NaCl-30+H2O-- ΔHneut =-57.1KJ

ΔH=-57.1×30×10-3×103J=1713 J

q=m·s.ΔT

1713=500×4.18×ΔT

ΔT=0.8196 K=81.96×10-2 K≈82×10-2 K

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (27 Aug Shift 1)
$200 \mathrm{ml}$ of a solution contains $5.85 \mathrm{~g}$ dissolved sodium chloride. The concentration of the solution will be $(\mathrm{Na}=23 ; \mathrm{Cl}=35.5)$

Options:
A 1 molar
B 2 molar
C 0.5 molar
D 0.25 molar

The correct answer is:
0.5 molar

Solution:
Molar concentration $=\frac{5.85 \times 1000}{58.5 \times 200}=0.5$ Molar

Topic: Chemistry, Solutions, JEE Main
$200 \mathrm{~mL}$ of an aqueous solution of a protein contains its $1.26 \mathrm{~g}$. The osmotic pressure of this solution at $300 \mathrm{~K}$ is found to be $2.57 \times 10^{-3}$ bar. The molar mass of protein will be

Options:
A $51022 \mathrm{~g} \mathrm{~mol}^{-1}$
B $122044 \mathrm{~g} \mathrm{~mol}^{-1}$
C $31011 \mathrm{~g} \mathrm{~mol}^{-1}$
D $61038 \mathrm{~g} \mathrm{~mol}^{-1}$

The correct answer is:
$61038 \mathrm{~g} \mathrm{~mol}^{-1}$

Solution:
$$
\begin{aligned}
& \because \pi=C R T \\
&=\frac{w \times 1000}{M \times V(\text { in mL })} \times R T \\
& \quad\left[\because C=\frac{n \times 1000}{V(\text { in mL })} \text { and } n=\frac{w}{M}\right] \\
&=\frac{1.26 \times 1000 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 200} \\
&=61038 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$$

Topic: Chemistry, Solutions, NEET, NEET 2011 (Mains)
245 g impure sample of KClO3s on heating gives 12 g  O 2 g according to 2  KClO 3 s →2 KCl s  + 3  O 2 g
Calculate % purity of sample ? If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
25

Solution:
2  KClO 3 →2 KCl+3  O 2
Let the % purity = x
Mole of KClO3 = 245122.5×x100 = 2x100
Mole of O2 = 2x100×32
2x100×32×32=12
(% purity of sample) x = 12.5

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
245 g impure sample of KClO3s on heating gives 12 g  O 2 g according to 2  KClO 3 s →2 KCl s  + 3  O 2 g
Calculate % purity of sample ?

Options:

The correct answer is:
12.5

Solution:
2  KClO 3 →2 KCl+3  O 2
Let the % purity = x
Mole of KClO3 = 245122.5×x100 = 2x100
Mole of O2 = 2x100×32
2x100×32×32=12
(% purity of sample) x = 12.5

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
250 mL of 0.5M NaOH was added to 500 mL of 1M HCl. The number of unreacted HCl molecules in the solution after the complete reaction is p×1021. Find out p
(Nearest integer) NA=6.022×1023

Options:

The correct answer is:
226

Solution:
We known that no. of moles =Vlitre × Molarity & No. of millimoles =Vml× Molarity so millimoles of NaOH=250×0.5
=125
Millimoles of HCl=500×1=500
Now reaction is
so millimoles of HCl left =375 Moles of HCl=375×10-3
No. of HCl molecules =6.022×1023×375×10-3
=225.8×1021
≃226×1021=226

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (20 Jul Shift 1)
250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO3 and 0.1M AuCl. The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electrodeposited will be : EAg+/Ag0=0.80V,EAu+/Au0=1.69V

Options:
A only gold
B silver and gold in proportion to their atomic weights
C only silver
D silver and gold in equal mass proportion

The correct answer is:
only gold

Solution:
Charge (q)= it 96500F=1×15×6096500=90096500=9965F=0.0093 F

No. of moles of Au+=0.025 & No. of moles of Ag+=0.025

Species with higher value of SRP will get deposited first at cathode.

(i) Au+(aqs) + e-⟶ Au(s)

0.025 0.0093 mole

so only Au will get deposited.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2020 (04 Sep Shift 2)
$25.3 \mathrm{~g}$ of sodium carbonate, $\mathrm{Na}_2 \mathrm{CO}_3$ is dissolved in enough water to make $250 \mathrm{~mL}$ of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, $\mathrm{Na}^{+}$and carbonate ion, $\mathrm{CO}_3^{2-}$ are respectively (Molar mass of $\mathrm{Na}_2 \mathrm{CO}_3=106 \mathrm{~g} \mathrm{~mol}^{-1}$ )

Options:
A $0.955 \mathrm{M}$ and $1.910 \mathrm{M}$
B $1.910 \mathrm{M}$ and $0.955 \mathrm{M}$
C $1.90 \mathrm{M}$ and $1.910 \mathrm{M}$
D $0.477 \mathrm{M}$ and $0.477 \mathrm{M}$

The correct answer is:
$1.910 \mathrm{M}$ and $0.955 \mathrm{M}$

Solution:
$$
\begin{aligned}
\text { Molarity } & =\frac{\text { number of moles of solute }}{\text { volume of solution (in mL) }} \times 1000 \\
& =\frac{25.3 \times 1000}{106 \times 250}=0.9547 \approx 0.955 \mathrm{M}
\end{aligned}
$$
$\mathrm{Na}_2 \mathrm{CO}_3$ in aqueous solution remains dissociated as
Since, the molarity of $\mathrm{Na}_2 \mathrm{CO}_3$ is $0.955 \mathrm{M}$, the molarity of $\mathrm{CO}_3^{2-}$ is also $0.955 \mathrm{M}$ and that of $\mathrm{Na}^{+}$is $2 \times 0.955=1.910 \mathrm{M}$

Topic: Chemistry, Solutions, NEET, NEET 2010 (Screening)
2.56×10-3 equivalent of KOH is required to neutralise 0.12544 g  H 2 XO 4 . The atomic mass of X (in g/mol ) is:
[Given: H 2 XO 4 is a dibasic acid]

Options:
A 16
B 8
C 7
D 32

The correct answer is:
32

Solution:
Let the atomic weight of X = y
No. of equivalents of KOH = No. of equivalents of H 2 XO 4
i.e. 0.12544 66 + y 2  = 0.12544
∴ y = 32

Topic: Chemistry, Redox Reactions, JEE Main
2.68Ã 10 â3 moles of a solution containing an ion A n+ require 1.61Ã 10 â1 moles of MnO 4 - ion for the oxidation of A n+ to AO 3 - in acidic medium. What is the value of n?

Options:

The correct answer is:
2.00

Solution:
A n+ â A +5 O 3 â + 5ân e â Ã 5, Oxidation half reaction
MnO4-+8H++5e-âMn2++4H2O Ã (5-n), Reduction half reaction
Applying reaction stoichiometry,
5 moles of An+ = 5-n moles of MnO 4-
5 5-n = 2.68 Ã 1 0 - 3 1.61 Ã 10 - 3 = 1.66 = 5 3
5 - n = 3
â´ n = 2

Topic: Chemistry, Redox Reactions, JEE Main
26.8 gm Na2SO4.nH2O contains 12.6 of water here the value of 'n' is?

Options:

The correct answer is:
7

Solution:
Na2SO4.nH2O
Molar mass of this salt =142+18n
Mass of water =12.626.8×142+18n
18n=12.626.8142+18n
On solving n=7

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$2.76 \mathrm{~g}$ of silver carbonate on being strongly heated yield a residue weighing

Options:
A $2.16 \mathrm{~g}$
B $2.48 \mathrm{~g}$
C $2.64 \mathrm{~g}$
D $2.32 \mathrm{~g}$

The correct answer is:
$2.16 \mathrm{~g}$

Solution:
$\begin{aligned} & 2 \mathrm{Ag}_2 \mathrm{CO}_3 \xrightarrow{\Delta} 4 \mathrm{Ag}+2 \mathrm{CO}_2+\mathrm{O}_2 \\ & 2 \times 276 \mathrm{gm} \quad 4 \times 108 \mathrm{gm}\end{aligned}$
■ $2 \times 276 \mathrm{gm}$ of $\mathrm{Ag}_2 \mathrm{CO}_3$ gives $4 \times 108 \mathrm{gm}$
$\therefore 1 \mathrm{gm}$ of $\mathrm{Ag}_2 \mathrm{CO}_3$ gives $=\frac{4 \times 108}{2 \times 276}$
$\therefore \quad 2.76 \mathrm{gm}$ of $\mathrm{Ag}_2 \mathrm{CO}_3$ gives
$\frac{4 \times 108 \times 2.76}{2 \times 276}=2.16 \mathrm{gm}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$2.76 \mathrm{~g}$ of silver carbonate on being strongly heated yield a residue weighing

Options:
A $2.16 \mathrm{~g}$
B $2.48 \mathrm{~g}$
C $2.64 \mathrm{~g}$
D $2.32 \mathrm{~g}$

The correct answer is:
$2.16 \mathrm{~g}$

Solution:
$2 \mathrm{Ag}_2 \mathrm{CO}_3 \xrightarrow{\Delta} 4 \mathrm{Ag}+2 \mathrm{CO}_2+\mathrm{O}_2$
$2 \times 276 \mathrm{gm} \quad 4 \times 108 \mathrm{gm}$
$\begin{aligned} & \text { Q } 2 \times 276 \mathrm{gm} \text { of } \mathrm{Ag}_2 \mathrm{CO}_3 \text { gives } 4 \times 108 \mathrm{gm} \\ & \therefore \quad 1 \mathrm{gm} \text { of } \mathrm{Ag}_2 \mathrm{CO}_3 \text { gives }=\frac{4 \times 108}{2 \times 276} \\ & \therefore 2.76 \mathrm{gm} \text { of } \mathrm{Ag}_2 \mathrm{CO}_3 \text { gives } \\ & \frac{4 \times 108 \times 2.76}{2 \times 276}=2.16 \mathrm{gm}\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
2.76 g of silver carbonate on being strongly heated yields a residue weighing

Options:
A 2.16 g
B 2.48 g
C 2.32 g
D 2.64 g

The correct answer is:
2.16 g

Solution:
Ag2CO32.76→ΔAg2O+CO2→Δ2Ag+12O2residue
Note: Ag2O is thermally unstable and decompose on heating liberating oxygen.
Molecular weight of Ag2CO3=108×2+12+16×3=276g
So 276 g of Ag2CO3 on heating gives residue
=2×108=216g of Ag
So 2.76 of Ag2CO3 on heating gives =216276×2.76
=2.16g of Ag

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
29.2 % w/w HCl stock solution has a density of 1.25 g mL-1 . The molecular weight of HCl is 36.5 g mol-1 . The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is

Options:

The correct answer is:
8

Solution:
Density of HCl=1.25 g mL solution -1
Weight of 100 mL solution = 125 g
For a 29.2 % w/w HCl solution, weight =125100Ã29.2=36.5 gm
So in 100 mL solution, moles of HCl =36.536.5=1 moles
So molarity of HCl=10 M
Using M1V1=M2V2
Volume required =V2Ã10=200Ã0.4
V2=8 mL

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$29.5 \mathrm{mg}$ of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absorbed in $20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}$ solution. The excess of the acid required $15 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$ solution for complete neutralization. The percentage of nitrogen in the compound is

Options:
A
$59.0$

B
$47.4$

C
$23.7$

D
$29.5$

The correct answer is:
$23.7$

Solution:
Moles of $\mathrm{HCl}$ reacting with ammonia
$$
\begin{aligned}
& =(\text { moles of } \mathrm{HCl} \text { absorbed })-(\text { moles of } \mathrm{NaOH} \text { solution required) } \\
& =\left(20 \times 0.1 \times 10^{-3}\right)-\left(15 \times 0.1 \times 10^{-3}\right) \\
& =\text { moles of } \mathrm{NH}_3 \text { evolved. } \\
& =\text { moles of nitrogen in organic compound }
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \text { wt. of nitrogen in org. comp }=0.5 \times 10^{-3} \times 14 \\
& =7 \times 10^{-3} \mathrm{~g} \\
& \% w t=\frac{7 \times 10^{-3}}{29.5 \times 10^{-3}}=23.7 \% \\
&
\end{aligned}
$$

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2010
29.5 mg of organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absorbed in 20mL of 0.1 M HCl solution. The excess of the acid required 15mL of 0.1M NaOH solution for complete neutralisation. The percentage of nitrogen in the compound is

Options:
A 59.0
B 47.4
C 23.7
D 29.5

The correct answer is:
23.7

Solution:
Weight of organic compound = 29.5 mg
NH 3 + HCl â NH 4 Cl HCl + NaOH â NaCl + H 2 O
(Remaining) 15Ã0.1 M
= 1.5 mmol
Total millimole of HCl = 2
Millimole used by NH3=2-1.5=0.5
Weight of NH3 = 0.5 Ã 17 mg = 8.5 mg
Weight of nitrogen =1417Ã8.5mg=7mg
% Nitrogen=729.5Ã100=23.7%

Topic: Chemistry, General Organic Chemistry, JEE Main
300 calories of heat is given to a heat engine, and it rejects 225 calories of heat. If source temperature is 227°C, then the temperature of sink will be _____ C∘.

Options:

The correct answer is:
102

Solution:
Efficiency of Carnot engine is given by η=1-Q2Q1=1-T2T1.

⇒1-225300=1-T2500

T2=53×225=375 K

So, temperature of sink is T2=375-273=102°C

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (29 Jun Shift 1)
$300 \mathrm{gm}$ of water at $25^{\circ} \mathrm{C}$ is added to $100 \mathrm{gm}$ of ice at $0^{\circ} \mathrm{C}$. The final temperature of the mixture is

Options:
A $12.5^{\circ} \mathrm{C}$
B $0^{\circ} \mathrm{C}$
C $25^{\circ} \mathrm{C}$
D $50^{\circ} \mathrm{C}$

The correct answer is:
$0^{\circ} \mathrm{C}$

Solution:
Ref. at $0^{\circ} \mathrm{C}$ water
$(\Delta \mathrm{H})$ liberated $=300 \times 1 \times 25=7500 \mathrm{cal}$
$(\Delta \mathrm{H})$ required $=\mathrm{m}_{\text {ice }} \mathrm{L}_{\mathrm{f}}$
$=100 \times 80 \mathrm{cal}=8000 \mathrm{cal}$
as $(\Delta \mathrm{H})_{\text {req. }}>(\Delta \mathrm{H})_{\text {iib }}$ the mixture remains at $0^{\circ} \mathrm{C}$
$\therefore$ Final temperature of mixture will be $0^{\circ} \mathrm{C}$.

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2021
$300 \mathrm{~J}$ of work is done in sliding a $2 \mathrm{~kg}$ block up an inclined plane of height 10 $\mathrm{m}$. Taking $g=10 \mathrm{~m} / \mathrm{s}^2$, work done against friction is:

Options:
A $1000 \mathrm{~J}$
B $200 \mathrm{~J}$
C $100 \mathrm{~J}$
D zero

The correct answer is:
$100 \mathrm{~J}$

Solution:
Potential energy
$$
\begin{aligned}
& =2 \times 10 \times 10 \\
& =200 \mathrm{~J}
\end{aligned}
$$
and work done $=300 \mathrm{~J}$
$\therefore$ Work done against friction
$$
=300-200=100 \mathrm{~J} .
$$

Topic: Physics, Work Power Energy, JEE Main
$450 \mathrm{mg}$ of glucose is dissolved in $100 \mathrm{~g}$ of solvent. What is the molality of solution?

Options:
A $0.0025 \mathrm{~m}$
B $0.025 \mathrm{~m}$
C $0.25 \mathrm{~m}$
D $2.5 \mathrm{~m}$

The correct answer is:
$0.025 \mathrm{~m}$

Solution:
Molality $=\frac{\text { moles of solute } \times 1000}{\text { weight of solvent (in g) }}$
Or $=\frac{\text { weight of solute } \times 1000}{\text { mol. wt of solute } \times \text { weight of solvent (in g) }}$
$$
=\frac{450 \times 10^{-3} \times 1000}{180 \times 100}
$$
$[\because$ Mol. mass of glucose $=180]$
$=0.025 \mathrm{~m}$

Topic: Chemistry, Solutions, JEE Main
500 g impure sample of CaCO 3 on heating gives 70âg of CaO. Percentage impurities in sample is

Options:
A 25%
B 50%
C 75%
D 80%

The correct answer is:
75%

Solution:
CaCO3→CaO+CO2
Let the % purity is x
Mol: 500×X100×100 mol
500×X100×100×56=70
X=25
% Impurities =100-25=75%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$500 \mathrm{~g}$ of water and $100 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$ are in a calorimeter whose water equivalent is $40 \mathrm{~g} .10 \mathrm{~g}$ of steam at $100^{\circ} \mathrm{C}$ is added to it. Then water in the calorimeter is : (Latent heat of ice $=80 \mathrm{cal} / \mathrm{g}$, Latent heat of steam $=540 \mathrm{cal} / \mathrm{g}$ )

Options:
A
$580 \mathrm{~g}$

B
$590 \mathrm{~g}$

C
$600 \mathrm{~g}$

D
$610 \mathrm{~g}$

The correct answer is:
$590 \mathrm{~g}$

Solution:
As $1 \mathrm{~g}$ of steam at $100^{\circ} \mathrm{C}$ melts $8 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$.
$10 \mathrm{~g}$ of steam will melt $8 \times 10 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$ Water in calorimeter $=500+80+10 \mathrm{~g}=590 \mathrm{~g}$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2013 (23 Apr Online)
500 mL of 0.150 M AgNO3 solution were added in 500 mL of 1.09 M Fe2+ solution and the reaction is allowed to reach an equilibrium at 2 5 ∘ C.

Ag + aq + Fe 2 + aq ⇌ Fe 3 + aq + Ag s

For 25 mL of the solution, 30 mL of 0.0832 M KMnO4 were required for oxidation. Calculate equilibrium constant (M-1) of the reaction at 2 5 ∘ C.
[Report your answer by rounding it upto nearset whole number]

Options:

The correct answer is:
3

Solution:
Ag + aq + Fe 2 + aq â Fe 3 + aq + Ag s Initial 7 5 mM 5 4 5 mM 0 0 Reaction - 4 6 mM - 4 6 mM + 4 6 mM + 4 6 mM --------------- --------------- --------------- --------------- Final 2 9 mM 4 9 9 mM 4 6 mM 4 6 mM
m.eqs of KMnO 4 = 3 0 Ã 0 Â· 0 8 3 2 Ã 5 = 1 2 Â· 4 8 m.eqs
(Z factor =5)
m.eqs of Fe 2 + = 12Â·48 Meqs = 12Â·48 mM
(Z factor= 1)
Total mM of Fe 2 + left = 12 Â· 48 Ã 4 0 = 4 9 9 mM
At Equilibrium concentrations
Ag + = 2 9 mM 1 0 0 0 mL = 0 Â· 0 2 9 M
Fe 2 + = 4 9 9 mM 1 0 0 0 mL = 0 Â· 4 9 9 M
Fe 3 + = 4 6 mM 1 0 0 0 mL = 0 Â· 0 4 6 M
K C = Fe 3+ Ag + Fe 2+ = 0Â·046 0Â·499Ã0Â·029 =3.18 M â1

Topic: Chemistry, Chemical Equilibrium, JEE Main
512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is V in Volt.

Options:

The correct answer is:
128

Solution:
Q=512 q

Volume i= Volume f

512×43πr3=43πR3

29r3=R3

R=8 r

2=kqr

V=kQR=k512q8r

V=128 V.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (25 Feb Shift 1)
$58.4 \mathrm{g}$ of $\mathrm{NaCl}$ and $180 \mathrm{g}$ of glucose were separately dissolved in $1000 \mathrm{mL}$ of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.

Options:
A NaCl solution will show higher elevation of boiling point
B Glucose solution will show higher elevation of boiling point
C Both the solutions will show equal elevation of boiling point
D The boiling point elevation will be shown by neither of the solutions

The correct answer is:
NaCl solution will show higher elevation of boiling point

Solution:
Elevation in boiling point, $\Delta T_{b}=i \times K_{b} \times m$
Molality of NaCl solution $=\frac{n}{W} \times 1000$
$=\frac{58.5}{W_{\mathrm{H}_{2} \mathrm{O}}} \times 1000=\frac{1000}{\mathrm{W}_{\mathrm{M}_{2}} \mathrm{O}}$
Molality of $f \mathrm{C}_{6} \mathrm{H}_{1} \not \mathrm{O}_{6}$ solution = $\frac{\frac{180}{180} \times 1000}{w_{\mathrm{H}_{2}} \mathrm{O}}$
$\frac{1000}{W_{M_{2}} O}$. Both the solutions have same molarity but the values for $\mathrm{NaCl}$ and glucose are ${2}$ and 1respectively. $\left.\therefore \quad \Delta T_{b(\mathrm{NaG})}=2 \times \Delta T_{b C C_{6} H_{1} 2} \mathrm{O}_{6}\right)$

Topic: Chemistry, Solutions, JEE Main