datavorous/entrance-exam-dataset huggingface dataset:
Sodium carbonate is prepared by Solvay process. Which of the following compounds is obtained as a by-product?

Options:
A NH4Cl
B NH3
C CO2
D CaCl2

The correct answer is:
CaCl2

Solution:
Here CaCl2 is formed as a by-product
NH3+H2O+CO2âNH4HCO3 â¦(i)
Na++Cl-+NH4++HCO3-âNaHCO3(ppt.)+NH4++Cl- â¦(ii)
For step (i), CO2 is produced by heating limestone.
CaCO3âCaO+CO2
CaO so-obtained is slaked with H2O and NH4Cl produced in step (ii) reacts to form CaCl2 as a by-product.
CaO +  H 2 O→ Ca(OH) 2 → NH 4 Cl CaCl 2  +  2NH 3  +  2H 2 O

Topic: Chemistry, s Block Elements, JEE Main
Sodium carbonate reacts with $\mathrm{SO}_2$ in aqueous medium to give

Options:
A $\mathrm{NaHSO}_3$
B $\mathrm{Na}_2 \mathrm{SO}_3$
C $\mathrm{NaHSO}_4$
D $\mathrm{Na}_2 \mathrm{SO}_4$

The correct answer is:
$\mathrm{NaHSO}_3$

Solution:
$\mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O}+2 \mathrm{SO}_2 \rightarrow 2 \mathrm{NaHSO}_3+\mathrm{CO}_2$

Topic: Chemistry, s Block Elements, JEE Main
Sodium carbonate, which is one of the most important products of the chemical industry, is prepared by the Solvay process based on the interaction of sodium chloride with ammonia and carbon dioxide. The reaction yields

Options:
A NH4HCO3
B NH4Cl
C NaHCO3
D (NH4)2CO3

The correct answer is:
NaHCO3

Solution:
NH3+H2O+CO2→NH4HCO3
Na++Cl-+NH4++HCO3-→NaHCO3ppt.+NH4++Cl-
2NaHCO3→ΔNa2CO3+CO2+H2O

Topic: Chemistry, s Block Elements, JEE Main
Sodium chloride is an ionic compound, whereas hydrogen chloride is mainly covalent because

Options:
A Sodium is less reactive
B Hydrogen is non - metal
C Hydrogen chloride is a gas
D Electronegativity difference in the case of hydrogen and chlorine is less than 2.1

The correct answer is:
Hydrogen is non - metal

Solution:
Hydrogen is non metal and non metal atoms form covalent bond
The nature of a bond formed between two elements mainly depends on the difference of electronegativity. If difference of electronegativity of two atoms is more than 1.90, they from an ionic bond. And when this difference is less than 1.9, they form a covalent bond.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
Sodium ethoxide has reacted with ethanoyl chloride. The compound that is produced in the above reaction is :

Options:
A
2-Butanone

B
Ethyl chloride

C
Ethyl ethanoate

D
Diethyl ether

The correct answer is:
Ethyl ethanoate

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2011
Sodium extract is heated with concentrated HNO3 before testing for halogens because :

Options:
A Ag2S and AgCN are soluble in acidic medium.
B Silver halides are totally insoluble in nitric acid.
C S2- and CN- , if present, are decomposed by conc. HNO3 and hence do not interfere in the test.
D Ag reacts faster with halides in acidic medium

The correct answer is:
S2- and CN- , if present, are decomposed by conc. HNO3 and hence do not interfere in the test.

Solution:
S2- and CN- ions if present are decomposed by conc. HNO3 and hence do not interfere in the test.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2016 (10 Apr Online)
Sodium fusion extract of an organic compound gives a blood red colouration with few drops of FeCl 3 solution. This indicates the presence of

Options:
A Nitrogen in the organic compound
B Sulphur in the organic compound
C Nitrogen and sulphur in the organic compound
D Sulphur and chlorine in the organic compound

The correct answer is:
Nitrogen and sulphur in the organic compound

Solution:
The presence of both nitrogen and sulphur in the lassaigne's test in an organic compound shows a blood red colouration with FeCl3 solution. This is due to the formation of NaCNS in the sodium fusion extract.

Topic: Chemistry, Practical Chemistry, JEE Main
Sodium has body centred packing. Distance between two nearest atoms is $3.7 Å$. The lattice parameter is :

Options:
A $8.6 Å$
B $6.8 Å$
C $4.3 Å$
D $3.0 Å$

The correct answer is:
$4.3 Å$

Solution:
$$
\begin{aligned}
\frac{\sqrt{3} \mathbf{a}}{2} & =3.7 Å \\
\Rightarrow \quad \mathbf{a} & =\frac{2 \times 3.7}{\sqrt{3}}=4.3 Å
\end{aligned}
$$

Topic: Chemistry, Solid State, NEET, NEET 2009 (Mains)
Sodium metal can be stored under

Options:
A Benzene
B Kerosene
C Alcohol
D Toluene

The correct answer is:
Kerosene

Solution:
The correct option is (2) Kerosene
Sodium is a highly reactive element. If it is kept in open it can explosively react with oxygen to catch fire. Hence to prevent accidental damage sodium is immersed in kerosene oil.

Topic: Chemistry, s Block Elements, JEE Main
Sodium metal crystallises in a body centred cubic lattice with unit cell edge length of 4 Å. The radius of sodium atom is ___×10-1Å. (Nearest integer)

Options:

The correct answer is:
17

Solution:
Edge length of the unit cell (a) = 4 x 10-8cm

Radius of sodium atom (r)=?

Since, sodium crystallises in body centered cubic arrangement.

The radius of the atom and the edge length of the cubic unit cell are related as follows,

For bcc structure, 4r=3a⇒r=34a

r=1.732×4×10-84r=1.732×10-8 cmr=17.3Å.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å . The radius of sodium atom is approximately:

Options:
A 0.93Â â«
B 1.86 â«
C 3.22Â â«
D 5.72Â â«

The correct answer is:
1.86 â«

Solution:
For B.C.C lattice,
4 r = 3 a

where 'r' is radius of atom and 'a' is edge length of unit cell
r = 3 4 a = 1.732 4 × 4.29
= 1.86 Å

Topic: Chemistry, Solid State, JEE Main, JEE Main 2015 (04 Apr)
Sodium metal is extracted by

Options:
A Electrolysis of aqueous solution of sodium chloride
B Electrolysis of fused sodium chloride
C Heating sodium oxide with carbon
D Heating sodium oxide with hydrogen

The correct answer is:
Electrolysis of fused sodium chloride

Solution:
Sodium is obtained by the electrolysis of fused sodium chloride.

Topic: Chemistry, s Block Elements, JEE Main
Sodium metal on dissolution in liquid ammonia gives a deep blue solution due to the formation of:

Options:
A Sodamide
B Sodium-ammonia complex
C Sodium ion-ammonia complex
D Ammoniated electrons

The correct answer is:
Ammoniated electrons

Solution:
Na+x+yNH3→[(Na(NH3)x)]++[(e(NH3)y)]−
Blue colour is due to the ammoniated electrons.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2019 (10 Jan Shift 2)
Sodium nitroprusside, when added to an alkaline solution of sulphide ions, produces purple colour ion due to the formation of

Options:
A $\mathrm{Na}\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right), \mathrm{NOS}\right]$
B $\mathrm{Na}_2\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_3 \mathrm{NOS}\right]$
C $\mathrm{Na}_3[\mathrm{Fe}(\mathrm{CN}), \mathrm{NOS}]$
D $\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_3 \mathrm{NOS}\right]$

The correct answer is:
$\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_3 \mathrm{NOS}\right]$

Solution:
$\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]+\mathrm{S}^{2-} \rightarrow \underset{\substack{\text { Sodimm thio-nithoprusside } \\ \text { (violet) }}}{\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]}$

Topic: Chemistry, Practical Chemistry, JEE Main
Sodium oxide reacts with water to produce sodium hydroxide. 20.0 g of sodium oxide is dissolved in 500 mL of water. Neglecting the change in volume, the concentration of the resulting NaOH solution is _________ ×10-1M. (Nearest integer)
[Atomic mass : Na=23.0, O=16.0, H=1.0]

Options:

The correct answer is:
13

Solution:
Na2O+H2O→2NaOH
According to reaction 2062 mole of Na2O gives 4062 mole of NaOH
so molarity of NaOH solution is =nNaOHVml×1000=4062×1000500=8062=1.29M≈1.3M

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Sodium phenoxide reacts with CO2 at 400 K and 4-7âatm pressure to give

Options:
A sodium salicylate
B salicylaldehyde
C catechol
D benzoic acid

The correct answer is:
sodium salicylate

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Sodium sulphite is used in preserving squashes and other mildly acidic foods due to

Options:
A Potassium salt has preservative action
B Potassium metabisulphite prevents oxidation
C Potassium metabisulphite is not influcenced by acid
D Sulphur dioxide and sulphurous acid formed kill bacteria and germs

The correct answer is:
Sulphur dioxide and sulphurous acid formed kill bacteria and germs

Solution:
K2S2O5→K2SO3+SO2
SO2+H2O→H2SO3
SO2 and sulphurous acid H2SO3 kill the bacteria and prevent the decay of food.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Sodium sulphite on heating with dilute $\mathrm{HCl}$ liberates a gas which

Options:
A Turns lead acetate paper black
B Turns acidified potassium dichromate paper green
C Burns with a blue flame
D Smells like vinegar

The correct answer is:
Turns acidified potassium dichromate paper green

Solution:
$\mathrm{Na}_2 \mathrm{~S}+$ dil. $2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{H}_2 \mathrm{~S}$
$\mathrm{H}_2 \mathrm{~S}+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \xrightarrow{\text { acide }} \mathrm{K}_2 \mathrm{SO}_4$ $+\underset{\text { (green) }}{\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3}+\mathrm{S}$

Topic: Chemistry, Practical Chemistry, JEE Main
Sodium thiosulphate is prepared by

Options:
A Reducing $\mathrm{Na}_2 \mathrm{SO}_4$ solution with $\mathrm{H}_2 \mathrm{~S}$
B Boiling $\mathrm{Na}_2 \mathrm{SO}_3$ solution with $\mathrm{S}$ in alkaline medium
C Neutralising $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_3$ solution with $\mathrm{NaOH}$
D Boiling $\mathrm{Na}_2 \mathrm{SO}_3$ solution with $S$ in acidic medium

The correct answer is:
Boiling $\mathrm{Na}_2 \mathrm{SO}_3$ solution with $S$ in acidic medium

Solution:
Sodium thiosulphate \(\left(\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3\right)\) is prepared by:
\(8 \mathrm{Na}_2 \mathrm{SO}_3+\mathrm{S}_8 \xrightarrow[\text { in acidic medium }]{\text { Boiling }} 8 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3\)

Topic: Chemistry, s Block Elements, JEE Main
Sodium thiosulphate is prepared by

Options:
A reducing \( \mathrm{Na}_{2} \mathrm{SO}_{4} \) solution with \( \mathrm{H}_{2} \mathrm{~S} \)
B boiling \( \mathrm{Na}_{2} \mathrm{SO}_{3} \) solution with \( \mathrm{S} \) in alkaline medium
C neutralising \( \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \) solution with \( \mathrm{NaOH} \)
D boiling \( \mathrm{Na}_{2} \mathrm{SO}_{3} \) solution with \( \mathrm{S} \) in acidic medium

The correct answer is:
boiling \( \mathrm{Na}_{2} \mathrm{SO}_{3} \) solution with \( \mathrm{S} \) in alkaline medium

Solution:
In the acidified medium Na2S2O3 is unstable

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Sodium thiosulphate is used in photography

Options:
A To convert metallic silver into silver salt
B $\mathrm{AgBr}$ grain is reduced to non-metallic silver
C To remove reduced silver
D To remove undecomposed $\mathrm{AgBr}$ in the form of $\mathrm{Na}_3\left[\mathrm{Ag}\left[\mathrm{S}_2 \mathrm{O}_3\right)_2\right]$ (a complex salt)

The correct answer is:
To convert metallic silver into silver salt

Solution:
Sodium thiosulphate is a reducing agent which convert metalic silver into silver salt.

Topic: Chemistry, s Block Elements, JEE Main
Sodium when heated in a current of dry ammonia gives

Options:
A Sodium nitrite
B Sodium hydride
C Sodium amide
D Sodium azide

The correct answer is:
Sodium amide

Solution:
$2 \mathrm{Na}+2 \mathrm{NH}_3 \longrightarrow 2 \mathrm{NaNH}_2+\mathrm{H}_2$

Topic: Chemistry, s Block Elements, JEE Main
Solar energy is mainly caused due to:

Options:
A burning of hydrogen in the oxygen
B fission of uranium present in the Sun
C fusion of protons during synthesis of heavier elements
D gravitational contraction

The correct answer is:
fusion of protons during synthesis of heavier elements

Solution:
Solar energy $\rightarrow$ fusion of protons into helium

Topic: Physics, Nuclear Physics, NEET, NEET 2003
Solenocytes are the main excretory structures in

Options:
A annelids
B molluses
C echinodermates
D platyhelminthes

The correct answer is:
platyhelminthes

Solution:
Solenocytes (also called flame cells) are meant for excretion and osmoregulation in Platyhelminthes. Annelids have metanephridia for excretion. Molluscs have kidneys (although different from vertebrate kidneys) for excretion.
In echinodermates, no special excretory organs are found for excretion. In them, excretion takes place through diffusion or osmosis or through active transport.

Topic: Biology, Animal Kingdom, JIPMER, JIPMER 2011
Solid Ba $\left(\mathrm{NO}_3\right)_2$ is gradually dissolved in a $1.0 \times 10^{-4} \mathrm{M} \mathrm{Na}_2 \mathrm{CO}_3$ solution. At what concentration of $\mathrm{Ba}^{2+}$ will a precipitate begin to form ? $\left(\mathrm{K}_{\mathrm{sp}}\right.$ for $\left.\mathrm{Ba} \mathrm{CO}_3=5.1 \times 10^{-9}\right)$.

Options:
A
$4.1 \times 10^{-5} \mathrm{M}$

B
$5.1 \times 10^{-5} \mathrm{M}$

C
$8.1 \times 10^{-8} \mathrm{M}$

D
$8.1 \times 10^{-7} \mathrm{M}$

The correct answer is:
$5.1 \times 10^{-5} \mathrm{M}$

Solution:
$$
\begin{aligned}
& \mathrm{Ba}\left(\mathrm{NO}_3\right)_2+\mathrm{CaCO}_3 \rightarrow \mathrm{BaCO}_3+2 \mathrm{NaNO}_3 \\
& \text { Here }\left[\mathrm{CO}_3^{-2}\right]=\left[\mathrm{Na}_2 \mathrm{CO}_3\right]=10^{-4} \mathrm{M} \\
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{+2}\right]\left[\mathrm{CO}_3^{-2}\right] \Rightarrow 5.1 \times 10^{-9}=\left[\mathrm{Ba}^{2+}\right]\left(10^{-4}\right) \Rightarrow\left[\mathrm{Ba}^{+2}\right]=5.1 \times 10^{-5}
\end{aligned}
$$
At this value, just precipitation starts.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2009
Solid $\mathrm{Ba}\left(\mathrm{NO}_3\right)_2$ is gradually dissolved in a $1.0 \times 10^{-4} \mathrm{~M} \mathrm{~Na}_2 \mathrm{CO}_3$ solution. At which concentration of $\mathrm{Ba}^{2+}$, precipitate of $\mathrm{BaCO}_3$ begins to form ? $\left(K_{s p}\right.$ for $\left.\mathrm{BaCO}_3=5.1 \times 10^{-9}\right)$

Options:
A
$5.1 \times 10^{-5} \mathrm{M}$

B
$7.1 \times 10^{-8} \mathrm{M}$

C
$4.1 \times 10^{-5} \mathrm{M}$

D
$8.1 \times 10^{-7} \mathrm{M}$

The correct answer is:
$5.1 \times 10^{-5} \mathrm{M}$

Solution:
Given $\mathrm{Na}_2 \mathrm{CO}_3=1.0 \times 10^{-4} \mathrm{M}$
$\begin{aligned}
& \therefore \quad\left[\mathrm{CO}_3^{--}\right]=1.0 \times 10^{-4} \mathrm{M} \\
& \text { i.e. } \quad \mathrm{s}=1.0 \times 10^{-4} \mathrm{M}
\end{aligned}$
At equilibrium
$\begin{aligned}
& {\left[\mathrm{Ba}^{++}\right]\left[\mathrm{CO}_3{ }^{--}\right]=\mathrm{K}_{\mathrm{sp}} \text { of } \mathrm{BaCO}_3} \\
& {\left[\mathrm{Ba}^{++}\right]=\frac{\mathrm{K}_{\mathrm{sp}}}{\left[\mathrm{CO}_3{ }^{--}\right]}=\frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}}}
\end{aligned}$
$=5.1 \times 10^{-5} \mathrm{M}$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2013 (09 Apr Online)
Solid $\mathrm{CO}_2$ is known as dry ice, because

Options:
A It melts at $0^{\circ} \mathrm{C}$
B It evaporates at $40^{\circ} \mathrm{C}$
C It evaporates at $-78^{\circ} \mathrm{C}$ without melting
D Its boiling point is more than $199^{\circ} \mathrm{C}$

The correct answer is:
It evaporates at $-78^{\circ} \mathrm{C}$ without melting

Solution:
Solid $\mathrm{CO}_2$ is knows as dry ice because it evaporates at $-78^{\circ} \mathrm{C}$ without changing in the liquid state.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Solid fuel used in rocket is a mixture of Fe2O3 and Al (in ratio 1 : 2). The heat evolved (kJ) per gram of the mixture is _____ (Nearest integer)

Given ΔHf0Al2O3=-1700 kJ mol-1

ΔHf0Fe2O3=-840 kJ mol-1

Molar mass of Fe, AI and O are 56, 27 and 16 g mol–1 respectively

Options:

The correct answer is:
4

Solution:
Molten iron can be prepared by alumino thermite process as shown below.

Fe2O3+2Al→2Fe+Al2O3

The iron oxide and aluminium metal ratio is 1:2 in the above reaction.

Now, the Enthalpy of the reaction is equal to the difference of enthalpy of formation of products to the enthalpy of formation of reactants.

ΔHreaction 0=-1700+(840)

Ratio of Fe2O3 and Al

= 160 : 54

= 2.96 in 214 gm mixture

∴ΔH°1gm mixture =-860214=-4 kJ/gram

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (11 Apr Shift 1)
Solid N 2 O 5 is-

Options:
A Ionic
B Covalent
C Coordinate covalent
D Metallic

The correct answer is:
Ionic

Solution:
Solid N 2 O 5 is ionic. X-ray diffraction shows that solid N 2 O 5 exists as NO 2 + NO 3 − (nitronium nitrate).

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Solid Na2SO4 is slowly added to a solution which is 0.020 M in BaNO32 and 0.020 M in PbNO32. Assume that there is no increase in volume on adding Na2SO4 . There preferential precipitation takes place. What is the concentration of Ba 2+ when PbSO 4 starts to precipitate?

KspBaSO4=1.0 Ã 10-10 and KspPbSO4=1.6 Ã 10-8

Options:
A 5.0 Ã 10-9 M
B 8.0 Ã 10-7 M
C 1.25 Ã 10-4M
D 1.95 Ã 10-8M

The correct answer is:
1.25 Ã 10-4M

Solution:
When BaSO4 begins to precipitate
SO42-=KspBaSO4Ba2+=1.0Ã10-100.020
=5.0 Ã 10-9 M
When PbSO4 begins to precipitate
SO42-=KspPbSO4Pb2+=1.6Ã10-80.020
=8.0 Ã 10-7 M
SO42-is less for BaSO4 precipitation so BaSO4precipitates first when PbSO4 begins to precipitate SO42-=8.0 Ã 10-7M at that point concentration of Ba2+=KspBaSO4SO42-
=1.0Ã10-108.0Ã10-7=1.25 Ã 10-4 M

Topic: Chemistry, Ionic Equilibrium, JEE Main
Solid $\mathrm{NaCl}$ is a bad conductor of electricity since

Options:
A In solid $\mathrm{NaCl}$ there are no ions
B Solid $\mathrm{NaCl}$ is covalent
C In solid $\mathrm{NaCl}$ there is no motion of ions
D In solid $\mathrm{NaCl}$ there are no electrons

The correct answer is:
In solid $\mathrm{NaCl}$ there is no motion of ions

Solution:
Solid $\mathrm{NaCl}$ is a bad conductor of electricity because solid $\mathrm{NaCl}$ there are no ions to move. Electrical conductivity: Conductivity is the measure of the ease at which an electric charge or heat can pass through a material.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Solid sodium chloride is bad conductor of electricity because

Options:
A It contains only molecules
B It does not possess ions
C The ions present in it are not free to move
D It does not contain free molecules

The correct answer is:
The ions present in it are not free to move

Solution:
The ions are not free to move in solid state and held up in lattice due to strong coulombic forces of attraction.

Topic: Chemistry, Electrochemistry, JEE Main
Solids are heated by the mode of

Options:
A conduction
B radiation
C convection
D None of these

The correct answer is:
conduction

Solution:
In conduction, the transfer of heat takes place from one part of the body to another without the actual movements of the particle from their equilibrium position. The transfer of heat from one part to another of solids takes place by conduction.

Topic: Physics, Thermal Properties of Matter, JEE Main
Solubility of $\mathrm{AgCl}$ will be minimum in

Options:
A $0.001 \mathrm{M} \mathrm{AgNO}_3$
B Pure water
C $0.01 \mathrm{M} \mathrm{CaCl}_2$
D $0.01 \mathrm{M} \mathrm{NaCl}$

The correct answer is:
$0.01 \mathrm{M} \mathrm{CaCl}_2$

Solution:
$0.01 \mathrm{M} ~\mathrm{CaCl}_2$ gives maximum $\mathrm{Cl}^{-}$ions to keep $\mathrm{K}_{s p}$ of $\mathrm{AgCl}$ constant, decrease in $\left[\mathrm{Ag}^{+}\right] \text { will be maximum. }$.

Topic: Chemistry, Ionic Equilibrium, JEE Main
Solubility of $\mathrm{Ca}(\mathrm{OH})_{2}$ is $s \mathrm{~mol} \mathrm{~L}^{-1}$. The solubility product $\left(K_{s p}\right)$ under the same condition is

Options:
A $4 s^{3}$
B $3 s^{4}$
C $4 s^{2}$
D $s^{3}$

The correct answer is:
$4 s^{3}$

Solution:
$\begin{aligned} \mathrm{Ca}(\mathrm{OH}) & \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-} \\ K_{s p} &=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2} \\ &=(s)(2 s)^{2}=4 s^{3} \end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
Solubility of calcium phosphate (molecular mass, M) in water is Wg per 100 mL at 25°C. Its solubility product at 25°C will be approximately.

Options:
A 107WM3
B 107WM5
C 103WM5
D 105WM5

The correct answer is:
107WM5

Solution:
Given the solubility of Ca3PO42 is Wg/100 mL. The solubility in mol/L is S=WM×1000100mol/L.

The equilibrium reaction of calcium phosphate is

Ca3PO42⇌3Ca2++2PO43-At equilibrium3S2S

Now, Ksp=Ca2+3PO43-2

⇒Ksp=3S32S2=108×S5

⇒Ksp=108×WM×105⇒Ksp=108×105×WM5

=1.08×107WM5

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Solubility of calcium phosphate (molecular mass, M) in water is Wg per 100 mL at 25°C. Its solubility product at 25°C will be approximately.

Options:
A 107WM3
B 107WM5
C 103WM5
D 105WM5

The correct answer is:
107WM5

Solution:
Given the solubility of Ca3PO42 is Wg/100 mL. The solubility in mol/L is S=WM×1000100mol/L.

The equilibrium reaction of calcium phosphate is

Ca3PO42⇌3Ca2++2PO43-At equilibrium3S2S

Now, Ksp=Ca2+3PO43-2

⇒Ksp=3S32S2=108×S5

⇒Ksp=108×WM×105⇒Ksp=108×105×WM5

=1.08×107WM5

Topic: Chemistry, Ionic Equilibrium, JEE Main
Solubility product constant $\left(K_{\mathrm{sp}}\right)$ of salts of types $M X, M X_2$ and $M_3 X$ at temperature ' $T$ ' are $4.0 \times 10^{-8}, 3.2 \times 10^{-14}$ and $2.7 \times 10^{-15}$, respectively. Solubilities $\left(\mathrm{mol} \mathrm{dm}^{-3}\right)$ of the salts at temperature ' $T$ ' are in the order

Options:
A
$M X>M X_2>M_3 X$

B
$M_3 X>M X_2>M X$

C
$M X_2>M_3 X < M X$
D
$M X>M_3 X>M X_2$

The correct answer is:
$M X>M_3 X>M X_2$

Solution:
$\begin{array}{lcl}\text { Salt } & \text { Solubility product } & \text { Solubility } \\ M X & S_1^2=4.0 \times 10^{-8} & S_1=2 \times 10^{-4} \\ M X_2 & 4 S_2^3=3.2 \times 10^{-14} & S_2=2 \times 10^{-5} \\ M_3 X & 27 S_3^4=2.7 \times 10^{-15} & S_3=1 \times 10^{-4}\end{array}$
Thus, solubility order $=M X>M_3 X>M X_2$.

Topic: Chemistry, Ionic Equilibrium, JEE Main
Solubility product of a sulphide $M S$ is $3 \times 10^{-25}$ and that of another sulphide $N S$ is $4 \times 10^{-40}$. In ammoniacal solution

Options:
A Only NS gets precipitated
B Only MS gets precipitated
C No sulphide precipitates
D Both sulphides precipitates

The correct answer is:
Both sulphides precipitates

Solution:
The concentration of $S^{2-}$ ions in group II is lowered by maintaining acidic medium in the presence of $\mathrm{NH}_4 \mathrm{Cl}$.
The ionization of $\mathrm{H}_2 \mathrm{~S}$ is supressed due to common ion effect. So the ionic product is less than solubility product

Topic: Chemistry, Ionic Equilibrium, JEE Main
Solubility product of $\mathrm{AgBr}$ is $4.9 \times 10^{-13}$. What is its solubility?

Options:
A $2.4 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}$
B $3.2 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}$
C $4.9 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}$
D $7.0 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}$

The correct answer is:
$7.0 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}$

Solution:
$\begin{aligned} & \mathrm{K}_{\mathrm{sp}}(\mathrm{AgBr})=\mathrm{S}^2 \\ & 4.9 \times 10^{-13}=\mathrm{S}^2 \\ & \mathrm{~S}=\sqrt{4.9 \times 10^{-13}} \\ & \mathrm{~S}=7.0 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main