Chemistry

Electrochemistry and Solutions

206 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 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.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

An experiment showed that a lead chloride solution is formed when 6.21 g of lead combines with 4.26 g of chlorine. What is the empirical formula of this chloride? 

[Pb = 207; Cl = 35.5]

  1. $PbCl _3$
  2. $PbCl _2$
  3. $PbCl _4$
  4. $PbCl$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
  Mass        Atomic weight    Relative no. of moles    Simplest ratio
Lead           6.21 g        207     6.21/207 = 0.03    0.03/0.03 = 1
Chlorine 4.26 g        35.5     4.26/35.5 = 0.12    0.12/0.03 = 4


Hence, empirical formula is $PbCl _4$. 

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

Common salt obtained from Clifton beach contained $60.75\%$ chlorine while $6.40$ g of a sample of common salt from Khewra mine contained $3.888$ g of chlorine. State the law illustrated by these chemical combinations.

  1. Law of reciprocal proportion

  2. Law of multiple proportion

  3. Law of constant composition

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

First case :


Common salt from Clifton beach contains $=$ $60.75\%$ $Cl _2$

$100$ g of salt $=60.75$ g of $Cl _2$

$1$ g of salt $=\dfrac {60.75}{100}=0.6075$ g of $Cl _2$

Second case :


$6.40$ g of $NaCl$ from Khewra mine $=3.888$ g of $Cl _2$

$1$ g of $NaCl$ from Khewra mine $=\dfrac {3.888}{6.40}=0.6075$ g of $Cl _2$

Thus, the weight of $Cl _2$ in $1$ g of salt in both the cases is same. Hence, the law of constant composition is verified.


Hence the correct option is C.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

Potassium combines with two isotopes of chlorine $(^{35} Cl\,\, and\,\, ^{37}Cl)$ respectively to form two samples of $KCl$ Their formation follows the law of:

  1. constant proportions

  2. multiple proportions

  3. reciprocal proportions

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
According to the Law of Definite Proportions, a chemical compound will always have exactly the same proportion of elements by mass.
This means that the elements that make up a compound will always have the same per cent composition by mass, regardless of the actual mass of the sample.
In this case, potassium chloride has a molar mass of 74.551 g/mol. The two elements that form potassium chloride are potassium, which has a molar mass of 39.0983 g/mol, and chlorine, which has a molar mass of 35.4527 g/mol.
This tells you that every mole of potassium chloride weighs 74.551 g, out of which 39.0983 g is potassium and 35.4527 g is chlorine.

Therefore, chlorine and potassium will always be in a ratio by mass of
$\dfrac{35.4527g}{39.0983g}$=0.90681
so the constant proportion is the right answer 

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

The % loss in mass after heating a pure sample of potassium chlorate (Mol. mass = 122.5) will be:

  1. 12.25

  2. 24.50

  3. 39.17

  4. 49.0
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$2KClO _3\rightarrow 2KCl+3O _2$
$2\times 122.5$ grams shows a wieght loss of  $3\times32 grams$
So 245 grams of $KClO _3$ is 100%
96 grams of $O _2$ is X%
$x=\frac{100\times 96}{245}=39.17$.
Hence option C is correct.

Multiple choice biology movement in and out of the cell transport of molecules osmosis and plasmolysis plant water relation

Decreasing order concentration of minerals inside cell is

  1. Ca$^{+2}$$-$K$^+$$-$Na$^+$.
  2. K$^+$$-$Ca$^{+2}$$-$Na$^+$.
  3. K$^+$$-$Na$^+$$-$Ca$^{+2}$.
  4. Na$^+$$-$K$^+$$-$Ca$^{+2}$.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Potassium ions are usually higher inside the cell than outside the cell, and its concentration inside the cell is about 10 to 30 fold higher than the other components, while sodium ion is higher outside the cell than inside the cell. 
So, the answer is C.
Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

An aqueous solution of volume 500 ml, when the reaction : $ 2Ag^+(aq) +Cu(s) \leftrightharpoons Cu^{2+}(aq) +2Ag(s) $ reached equilibrium, the concentration of $ Cu^{2+} $ ions was xM. to this solution, 500 ml of water is added.at new equilibrium , the concentration of $ Cu^{2+} $ ions would be 

  1. 2x M

  2. x M

  3. between x and 0.5 x M

  4. less than 0.5 x M
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Adding water increases the volume, which shifts the equilibrium. According to Le Chatelier's principle, the system will try to increase the number of ions. However, the dilution effect (concentration = moles/volume) dominates, leading to a decrease in concentration.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

$NaCl$ is doped with $2\times { 10 }^{ -3 }$ mole % $Sr{Cl} _{2}$, the concentration of cation vacancies is

  1. $6.02\times { 10 }^{ 18 }{ mol }^{ -1 }$
  2. $1.204\times { 10 }^{ 19 }{ mol }^{ -1 }$
  3. $3.01\times { 10 }^{ 18 }{ mol }^{ -1 }$
  4. $1.204\times { 10 }^{ 21 }{ mol }^{ -1 }\quad $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Each Sr2+ ion replaces two Na+ ions in the lattice, creating one cation vacancy to maintain charge neutrality. 2 * 10^-3 mole % means 2 * 10^-5 moles of SrCl2 per mole of NaCl. This corresponds to 2 * 10^-5 * 6.022 * 10^23 vacancies per mole, which is 1.204 * 10^19.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

A big irregular shaped vessel contained water, conductivity of which was $ 2.56 \times 10^{-3}\, S^{-1} \, m^{-1}.$ 585 g of NaCI was then added to the water and conductivity after the addition of Nacl, was found to be $ 3.06 \times 10^{-3}\,S^{-1}\, m^{-1}. $ The molar conductivity of Nacl at this concentration is $ 1.5 \times 10^{-2}\, S^{-1}\, mol^{-1}.$The capacity of vessel if it is fulfilled with water, is

  1. $ 3 \times 10^{4} 1$
  2. $ 30 \,1$
  3. $ 3 \times 10^{8} 1$
  4. $ 3 \times 10^{5} 1 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis
Aqueous solution of nickel sulphate contains $Ni^{2+}$ and ${SO _{4}}^{2-}$ ions. What will be the product at the nickel anode?
  1. $Ni^{2+}$
  2. ${SO _{4}}^{2-}$
  3. $Ni$
  4. $H _2SO _4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At the anode, oxidation occurs. Nickel metal (the anode) loses electrons to form Ni2+ ions, which dissolve into the solution.