Chemistry

Electrochemistry and Solutions

205 Questions

Work through advanced chemistry questions focusing on electrochemistry and properties of solutions. Key themes include electrolytic conductivity, colloidal solutions, and solubility products. These concepts are frequently tested in JEE advanced, NEET, and university level chemistry competitive examinations.

Colloidal solutionsMolar conductivitySolubility productElectrolytic conductivityHydrolysis of saltsEquivalent conductivity

Electrochemistry and Solutions Questions

Multiple choice conductivity and its types electrochemistry

The equivalent conductivity of monobasic acid at infinite dilution is 348 $ohm^{-1}$ $cm^2$ $eq^{-1}$. If the resistivity of the solution containing 15 g acid (molar mass 49) in 1 litre is 18.5 ohm cm, what is the degree of dissociation of acid?

  1. 45.9%

  2. 40.2%

  3. 60.4%

  4. 50.7%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

equivalent conductivity of monobar'c acid at infinite dilution $\wedge^{\circ} m = 348 \Omega^{-1} cm^2 eq^{-1}$

Amount of acid $= 15 g$
Molar mass $= 49$
Molarity = $\dfrac{15}{49} / 1 \, litre = 0.306 M$
Resistivity = $18.5$ ohm cm
conductivity = $\dfrac{1}{18.5} = 0.054$
Molar conductivity = $\dfrac{0.054}{0.306 \times 10^{-3}}$
$\wedge _m  = 176.64$
dissociation constant $\alpha = \dfrac{\wedge _m}{\wedge _m^{\circ}} = \dfrac{176.64}{348}$
$= 0.507$
$50.7 \%$
option $D$

Multiple choice conductivity and its types electrochemistry

The conductivities at infinite dilution of ${\text{N}}{{\text{H}} _{\text{4}}}{\text{Cl,NaOH}}$ and $\text{NaCl}$ are 130, 218, 120 ${\text{oh}}{{\text{m}}^{{\text{ - 1}}}}{\text{c}}{{\text{m}}^{\text{2}}}{\text{e}}{{\text{q}}^{{\text{ - 1}}}}$. If equivalent conductance of N/100 solution of ${\text{N}}{{\text{H}} _{\text{4}}}{\text{OH}}$ is 10, then degree of dissociation of ${\text{N}}{{\text{H}} _{\text{4}}}{\text{OH}}$ at this dilution is:

  1. 0.005

  2. 0.043

  3. 0.01

  4. 0.02
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$NH _4Cl+NaOH\longrightarrow NH _4OH+NaCl$


$\therefore \wedge _m^{\infty}$ $ _{NH _4OH}=\wedge _m^{\infty}$ 

$ _{NH _4Cl}+\wedge _m^{\infty}$ $NaOH-\wedge _m^{\infty}$ $ _{NaCl}$

$\wedge _m^{\infty}=130+218-120$
$\implies \wedge _m^{\infty}=228$ $scm^2eq^2$

$\alpha=\cfrac{\wedge _m}{\wedge _{m^{\infty}}}=\cfrac{10}{228}=0.0438$

Multiple choice chemistry nature of matter heterogenous mixture - colloidal solution and its properties suspensions and colloids mixtures

Which of the following will form negatively charged colloidal solution?

  1. $100$ ml $0.1$M $AgNO _3+100$ ml $0.1$ M KI
  2. $100$ ml $0.2$M $AgNO _3+100$ ml $0.1$M KI
  3. $100$ml $0.1$M $AgNO _3+100$ ml $0.2$ M KI
  4. $100$ ml $0.2$M $AgNO _3+200$ ml $0.1$M KI
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

milli moles of $AgNO _3=100\times0.1=10$

milli moles of $KI=100\times.02=20$

$KI$ is in excess.

$AgI$ will be precipitated and excess $KI$ as $I^-$ ions will surround $AgI$ and form negative charged colloid.


Therefore, correct option is C.

Multiple choice chemistry separation of substances classification of mixtures mixtures: examples and properties types of solutions

The amountof $H _2S$ required to precipitate $1.69$g $BaS$ from $BaCl _2$ solution is : 

  1. $3.4g$
  2. $0.24g$
  3. $0.34g$
  4. $0.17g$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$BaCl _2+H _2S\longrightarrow BaS+2HCl$
                                  $1.69$ $g$
For $1$ $mole$ of $H _2S$ produce $1$ $mole$ of $BaS$
So, $1$ $mole$ $H _2S\longrightarrow 1$ $mole$ $BaS$
            $34$ $g$                      $169$ $g$
                      $(x)\longleftarrow 1.69$ $g$
$\implies x=\cfrac {1.69\times 34}{169}=0.34$ $g$
Amount of $H _2S$ needed $=0.34$ $g$

Multiple choice chemistry the s-block elements (alkali and alkaline earth metals) anomalous properties of lithium group 1 elements: alkali metals properties of s block elements

There is loss in weight when mixture of $Li _{2}CO _{3}$ and $Na _{2}CO _{3}\cdot10H _{2}O$ is heate strongly. This loss is due to :

  1. $Li _{2}CO _{3}$
  2. $Na _{2}CO _{3}\cdot10H _{2}O$

  3. both (a) and (b)

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Loss in weight due to following reactions:
$Li _{2}CO _{3} \rightarrow Li _{2}O+CO _{2} \uparrow$
$Na _{2}CO _{3}\cdot10H _{2}O \rightarrow Na _{2}CO _{3} + 10H _{2}O \uparrow$
As sodium carbonate is thermally stable so it will not dissociates on heating.

Multiple choice chemistry changes - physical and chemical physical changes changes and their classification physical changes

Maximum conductivity would be of : 

  1. $K _{3}Fe(CN) _{6}[0.1 M Solution]$
  2. $K _{3}Ni(CN) _{6}[0.1 M Solution]$
  3. $FeSO _{4}.Al _{2}(SO _{4}) _{3}.24H _{2}O[0.1 M Solution]$
  4. $Na _{3}[Ag(S _{2}O _{3}) _{2}][0.1 M Solution]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Conductivity depends on the number of ions produced in solution. FeSO4.Al2(SO4)3.24H2O (Alum) dissociates into the most ions per formula unit compared to the complex salts listed, leading to higher molar conductivity.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

$CaCO _3 + 2HCl \rightarrow CaCl _2 + H _2O + CO _2$

The mass of calcium chloride formed when 2.5 g of calcium carbonate is dissolved in excess of hydrochloric acid is:

  1. 1.39 g

  2. 2.78 g

  3. 5.18 g

  4. 17.8 g
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

CaCO3 + 2HCl -----> CaCl2 + H2O + CO2. 
2.5g? g 
Mass of CaCO3= 40+12+3(16)=52+ 48= 100. 
Molar mass of CaCO3=100g. 
Mass of CaCl2= 40 + 2(35.5)= 40+ 71= 111. 
Molar mass of CaCl2=111g. 
Mass of CaCO3 (g) Mass of CaCl2 (g) 
For 100g of $CaCO _3$, 111g of CaCl_2$ is formed.
let for 2.5g of $CaCO_3$, $x$ g of $CaCl_2$ is formed.
Thus, by cross multiplication,

$x=111\times 2.5/100= 2.775g = 2.78g$.

Multiple choice chemistry hard water and soft water heavy water study of heavy water hydrogen and its compounds

Of the two solvent ${ H } _{ 2 }O$ and ${ D } _{ 2 }O$, $NaCl$ dissolves:

  1. equally in both solvents
  2. only in ${ H } _{ 2 }O$ but remains insoluble in ${ D } _{ 2 }O$
  3. more in ${ D } _{ 2 }O$
  4. more in ${ H } _{ 2 }O$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$NaCl$ (and other ionic compounds) dissolve more in $H _2O$ than in $D _2O$ as the dielectric constant of $H _2O$ $(78.39)$ is higher than the dielectric constant of $D _2O:(78.06)$. This is due to lower molar mass of $H _2O$ as compared to that of $D _2O$.

Multiple choice chemistry hard water and soft water heavy water study of heavy water hydrogen and its compounds

Which one of the following statements is correct about $D _{2}O$ and $H _{2}O$?

  1. $D _{2}O$ has lower dielectric constant than $H _{2}O$
  2. $NaCl$ is more soluble in $D _{2}O$ than in $H _{2}O$

  3. Both (a) and (b) are correct

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

D2O has a lower dielectric constant than H2O, which affects the solubility of ionic compounds. Consequently, many ionic salts, including NaCl, are less soluble in D2O than in H2O.

Multiple choice chemistry reactivity series and electrochemistry reactivity series and displacement reactions chemical properties of metals chemical properties of metals and non metals

A gas $Cl _{2}$ at $1\ atm$ is bubbled through a solution containing a mixture of $1\ M\ Br^{-1}$ and $1\ M\ F^{-1}$ at $25^{\circ}C$. If the reduction potential is $F > Cl > Br$, then

  1. $Cl$ will oxidise $Br$ and not $F$
  2. $Cl$ will oxidise $F$ and not $Br$
  3. $Cl$ will oxidise both $Br$ and $F$
  4. $Cl$ will reduce both $Br$ and $F$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Based on the reduction potential order F > Cl > Br, chlorine is a stronger oxidizing agent than bromine but weaker than fluorine. Therefore, chlorine will oxidize bromide ions (Br-) to bromine but cannot oxidize fluoride ions (F-).

Multiple choice chemistry water drying and dehydrating agents hydrolysis of salts and the ph of their solutions chemical equilibrium and acids-bases

The $K _{sp}$ of $Ag _{2}CrO _{4}, AgCl, AgBr$ and $AgI$ are respectively, $1.1\times 10^{-12}$, $1.8\times 10^{-10}$, $5.0\times 10^{-13}$ and $8.3\times 10^{-17}$. Which of the following salts will precipitate last if $AgNO _{3}$ solution is added to the solution containing equal moles of $NaCl, NaBr, NaI$ and $Na _{2}CrO _{4}$?

  1. $Ag _{2}CrO _{4}$
  2. $AgI$
  3. $AgCl$
  4. $AgBr$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
  1. $Ag _2CrO _4\rightleftharpoons 2Ag^++{CrO _4}^{2-}$

    $Ksp=(2s)^2\times s=4s^2$

    $Ksp=(1.1\times 10^{-12})$

    $S=3\sqrt {\cfrac {Ksp}{4}}=6.5\times 10^{-5}$M

    2. $AgCl\rightleftharpoons Ag^++Cl^-$

    $Ksp=S\times S$  ;       $Ksp=1.8\times 10^{-10}$

    $S=\sqrt {Ksp}=1.34\times 10^{-5}$M

    3. $AgBr\rightleftharpoons Ag^++Br^-$

    $Ksp=S\times S$ ;     $Ksp=5\times 10^{-13}$

    $S=\sqrt {Ksp}=0.71 \times 10^{-6}$M

    4. $AgI\rightleftharpoons Ag^++I^-$

    $Ksp=S\times S$  ;    $Ksp=8.3\times 10^{-17}$

    $S=\sqrt{Ksp}=0.9\times 10^{-8}$M

    $\therefore$ Solubility of $Ag _2CrO _4$ is highest, so it will precipitate last.
Multiple choice chemistry water drying and dehydrating agents hydrolysis of salts and the ph of their solutions chemical equilibrium and acids-bases

$NaCl + H _2SO _4\overset{\Delta}{\longrightarrow}HCl \ +$

  1. $NaHSO _4$
  2. $Na _2SO4$
  3. $Na _2SO _3$

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Reacting sodium chloride and sulphuric acid at elevated temperature leads to production of sodium bisulphate and hydrogen chloride. Little amount of sodium sulphate is also formed which reacts with the sulphuric acid to yield sodium bisulphate or sodium hydrogen sulphate.
$NaCl\left( s \right) +{ H } _{ 2 }{ SO } _{ 4 }\rightarrow { NaHSO } _{ 4 }\left( s \right) +HCl$

Multiple choice chemistry water drying and dehydrating agents hydrolysis of salts and the ph of their solutions chemical equilibrium and acids-bases

The hydrated salt $Na _{2}SO _{4}, 10H _{2}O$ undergoes $X\%$ loss in weight on heating and becomes anhydrous. The value of $X$ will be:

  1. $10$
  2. $45$
  3. $56$
  4. $70$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Molecular mass of $Na _2SO _410H _2O=2\times23+32+16\times14+1\times20=322$


On heating the $Na _2SO _410H _2O$ it becomes anhydrous that means 10 $H _2O$ molecules separated

Molecular mass of $10H _2O=20\times1+10\times16=180$

% loss $=\dfrac{180}{322}\times100=55.9$ % $ =56$ %

Multiple choice chemistry water drying and dehydrating agents hydrolysis of salts and the ph of their solutions chemical equilibrium and acids-bases

Which of the following salts will have maximum cooling effect when $0.5$ mole of the salt is dissolved in same amount of water. Integral heat of solution at $298\ K$ is given for each salt.

  1. $KNO _{3} (\Delta =35.4\ kJ\ mol^{-1})$
  2. $NaCl (\Delta =5.35\ kJ\ mol^{-1})$
  3. $HBr (\Delta =83.3\ kJ\ mol^{-1})$
  4. $KOH (\Delta =55.6\ kJ\ mol^{-1})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

More the heat absorbed, more will be the cooling effect. Hence more the positive value of $\Delta H$ more the cooling effect.In the given question KBr has $\Delta H =83.3 kJmol^-$ which is highest among all the gases.