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 metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

At infinite dilution equivalent conductances of $B{ a }^{ +2 }$ & $C{ 1 }^{ - }$ ions are $127$ & $76\ oh{ m }^{ -1 }c{ m }^{ -1 }\ e{ q }^{ -1 }$ respectively. Equivalent conductance of $BaCl _{ 2 }$ at infinite dilution is:

  1. $139.5$
  2. $101.5$
  3. $203$
  4. $279$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

${ \Lambda  } _{ eq }^{ \infty  }\left( { Ba }^{ 2+ } \right) =127{ \Omega  }^{ -1 }{ cm }^{ -1 }{ eq }^{ -1 }\ { \Lambda  } _{ eq }^{ \infty  }\left( { Cl }^{ - } \right) =76{ \Omega  }^{ -1 }{ cm }^{ -1 }{ eq }^{ -1 }\ { \Lambda  } _{ eq }^{ \infty  }\left( { Ba }{ Cl } _{ 2 } \right) ={ \Lambda  } _{ eq }^{ \infty  }\left( { Ba }^{ 2+ } \right) +2{ \Lambda  } _{ eq }^{ \infty  }\left( { Cl }^{ - } \right) \ \qquad \qquad \quad =127+76\times 2\ \qquad \qquad \quad =279{ \Omega  }^{ -1 }{ cm }^{ -1 }{ eq }^{ -1 }$

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

If equivalent conductance of 1M benzoic acid is $12.8\ { ohm }^{ -1 }{ cm }^{ 2 }$ and if the conductance of benzoate ion and ${ H }^{ -1 }$ ion are 42 and $288.42\ { ohm }^{ -1 }{ cm }^{ 2 }$ respectively. Its degree of dissolution is:

  1. 39%

  2. 3.9%

  3. 0.35%

  4. 0.039%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Lambda ^{ 0 } _{ m\left( C _{ 6 }H _{ 5 }COOH \right)  }={ \Lambda  } _{ (C _{ 6 }{ H } _{ 5 }CO{ O }^{ - }) }^{ 0 }\quad +{ \Lambda  }^{ 0 } _{ \left( { H }^{ + } \right)  }$ 

$\ \quad \quad \quad \quad \quad \quad \quad \quad =42+288.42=330.42\ now\quad we\quad have,\ \alpha =\frac { { \Lambda  } _{ m }^{ c } }{ { \Lambda  } _{ m }^{ 0 } } =\frac { 12.8 }{ 330.42 } =3.9$

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

The molar conductivity of cation and anion of salt $BA$ are $180$ and $220\ mhos\ cm^{2} mol^{-2}$ respectively. The molar conductivity of $BA$ at infinite dilution is:

  1. $90\ mhos \ cm^{2} mol^{-1}$
  2. $110\ mhos \ cm^{2} mol^{-1}$
  3. $400\ mhos \ cm^{2} mol^{-1}$
  4. $200\ mhos \ cm^{2} mol^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Molar conductivity at infinite dilution is the sum of the limiting molar conductivities of the cation and anion. For salt BA, we add the cation conductivity (180) and anion conductivity (220) to get 400 mhos cm² mol⁻¹. This additive property holds because at infinite dilution, ions behave independently without interionic interactions. The unit should be mol⁻¹, not mol⁻² as written in the question (a minor typographical error that doesn't affect the calculation).

Multiple choice imperfections in solids solid state the solid state chemistry

If 1 mole of NaCl is doped with $10^{-3}$ mole of $SrCl _2$. What is the number of cationic  vacancies per mole of NaCl ?

  1. $10^{-3}\, mole ^{-1}$
  2. $6.02\times 10^{18}\, mole ^{-1}$
  3. $10^{50}\, mole ^{-1}$
  4. $6.02\times 10^{20}\, mole ^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice imperfections in solids solid state the solid state chemistry

If $NaCl$ is doped with $10^{-3}$ mol$\%$ of $SrCl _2$, the concentration of cation vacancies will be: 

$(N _A=6.02\times 10^{23}mol^{-1}$)

  1. $6.02\times 10^{15} mol^{-1}$
  2. $6.02\times 10^{16} mol^{-1}$
  3. $6.02\times 10^{18} mol^{-1}$
  4. $6.02\times 10^{14} mol^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that $1$ $mol$ of $NaCl$ is doped with $\cfrac{10^{-3}}{100}$ $mol$ of $Sr^{+2}=10^{-5}$ $mol$


Cation vacancies produced by $Sr^{2+}$ ion $=1$               [$\because$ 1 $Sr^{+2}$ can replace 2 $Na^+$]


So, concentration of cation vacancies produced by $10^{-5}$ mole of $SrCl _2$

$=6.023\times 10^{23}\times 10^{-5}$

$=6.023\times 10^{18}$ per mole

Multiple choice imperfections in solids solid state the solid state chemistry

If $NaCl$ is doped with ${ 10 }^{ -4 }mol$% of ${ SrCl } _{ 2 }$, the concentration of cation vacancies will be: $\left( { N } _{ A }=6.02\times { 10 }^{ 23 }{ mol }^{ -1 } \right) $:

  1. $6.02\times { 10 }^{ 15 }{ mol }^{ -1 }$
  2. $6.02\times { 10 }^{ 16 }{ mol }^{ -1 }$
  3. $6.02\times { 10 }^{ 17 }{ mol }^{ -1 }$
  4. $6.02\times { 10 }^{ 14 }{ mol }^{ -1 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$NaCl$ is doped with $10^{-4}$ $mol\%$ of $SrCl _2$, i.e., one mole of $NaCl$ will have $10^{-6}$ mol of $SrCl _2$

$\rightarrow Sr^{+2}$ will replace one cation.
Therefore, concentration of cation vacancy $=10^{-6}\times 6.022\times 10^{23}=6.022\times 10^{17}$ $mol^{-1}$

Multiple choice imperfections in solids solid state the solid state chemistry

When NaCl is doped with ${ 10 }^{ -3 }$ mole % of Sr${ Cl } _{ 2 }$, what is the number of cationic vacancies?

  1. ${ 10 }^{ -5 }\times { N } _{ A }$
  2. ${ 10 }^{ -7 }\times { N } _{ A }$
  3. 6.022$\times{ 10 }^{8 }{ N } _{ A }$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice imperfections in solids solid state the solid state chemistry

If $100$ moles of NaCI are doped with ${10^{ - 3}}$ moles of $Sr{C _2}$ what is the concentration of cation vacancies?

  1. $6.02 \times {10^{18}}mo{l^{ - 1}}$
  2. $12.04 \times {10^{18}}mo{l^{ - 1}}$
  3. $3.01 \times {10^{18}}mo{l^{ - 1}}$
  4. $12.04 \times {10^{20}}mo{l^{ - 1}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

10^-3 moles of SrCl2 in 100 moles of NaCl means 10^-5 moles of SrCl2 per mole of NaCl. This creates 10^-5 * 6.02 x 10^23 = 6.02 x 10^18 vacancies per mole.

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

One litre of a sample of hard water contain $4.44mg$ $Ca{Cl} _{2}$ and $1.9mg$ of $Mg{Cl} _{2}$, what is the total hardness in terms of ppm of $Ca{CO} _{3}$ :

  1. $2$ ppm
  2. $3$ ppm
  3. $4$ ppm
  4. $6$ ppm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$1$ mole $CaCl _2\equiv 1$ mole $CaCO _3\equiv 1$ mole $MgCl _2$

$\therefore 100g$ $CaCO _3$ produces $111g$ of $CaCl _2$
$\therefore 4.44mg$ $CaCl _2$ produces $\cfrac {4.44\times 100}{111}mg$ $CaCO _3$
Similarly, $100g$ $CaCO _3$ is required for $95g$ $MgCl _2$
$\therefore 1.9mg$ $MgCl _2=\cfrac {1.9\times 100}{95}mg$ $CaCO _3$
                               $=2mg$ $CaCO _3$
Total hardness= Hardness due to $CaCl _2$+Hardness due to $MgCl _2$
                         =$4+2$
Total hardness=$6$ $ppm$

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

Unstable hardness of water is due to the presence of:

  1. $CaCl _{2},\:MgSO _{4}$
  2. $Ca^{+2},\:Mg^{+2}$
  3. $K^{+},\:CaCO _{3}$
  4. $Ca(HCO _{3}) _{2},\:Mg(HCO _{3}) _{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Hard water, water that contains salts of calcium and magnesium principally as bicarbonates, chlorides, and sulfates. Water hardness that is caused by calcium bicarbonate is known as temporary, because boiling converts the bicarbonate to the insoluble carbonate; hardness from the other salts is called permanent.

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

$Na _2CO _3$ is widely used in softening of hard water. If 1 L of hard water required $0.0106 g$ of $Na _2CO _3$, The hardness in ppm (parts per million i.e., $10^{6}$ ml) of $CaCO _3$ is:

  1. $0.01\,$ ppm $CaCO _3$
  2. $0.10\,$ ppm $CaCO _3$
  3. $1.00\,$ ppm $CaCO _3$
  4. $10.00\,$ ppm $CaCO _3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Hardness (in ppm)=$\cfrac {Weight\quad of\quad Na _2CO _3\quad required\quad (in\quad mg)}{Volume\quad of\quad Hard\quad water\quad (in\quad L)}$

=$\cfrac {0.0106\times 10^{3}}{1}$
=$10.6$
$\approx 10 ppm$
$\therefore$ Hardness (in ppm)= $10$ ppm $CaCO _3$

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

$ RH _{2} $ (ion exchange resin ) can replace $ Ca^{2+} $ in hard water as : 
$ RH _{2}+Ca^{2+}\rightarrow RCa+2H^{+} $.


One litre of hard water after passing through $ RH _{2} $ has pH = 2. Hence, hardness in ppm of $ Ca^{2+} $ is:

  1. 200

  2. 100

  3. 50

  4. 125
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to the given reaction 


$ RH _{2}+Ca^{2+}\rightarrow RCa+2H^{+} $

Each mole $ Ca^{2+} $ ion replaced by 2 moles $ H^{+}$

1 mole $ H^{+} $ replaced $ \Rightarrow \dfrac{1}{2} = 0.5\,mole \,Ca^{2+} $

Given,
$ pH = 2 $

$ H^{+} = 10^{-2} = 0.01 $

0.01 mole $ H^{+} $ replaced $ = 0.01\times 05 = 0.005\,moles\,Ca^{2+} $

Mass $ Ca^{2+}$ replaced $ = 0.005\times 40 = 0.2\,g = 200\,mg $

Concentration or Hardness of $ Ca^{2+} = 200\,mg/L $

$ = 200\,ppm $   

Hence, the correct option is $\text{A}$

Multiple choice evs experiments with water water and its types hard and soft water what floats - what sinks

Calculate the temporary and permanent hardness of water sample having the following the following constituents per litre:
$ Ca(HCO _{3}) _{2} = 162\, mg, MgCl _{2} = 95 =\, mg, $
$ NaCl = 585\, mg, Mg(HCO _{3}) _{2} = 73\, mg, $
$ CaSO _{4} = 136\, mg $ 

  1. 200 ppm, 150 ppm

  2. 100 ppm, 150 ppm

  3. 150 ppm, 200 ppm

  4. 150 ppm, 150 ppm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

mole of $ Ca(HCO _{3}) _{2} = \dfrac{162\times 10^{-3}mg}{162\,g/mole} = 1\times 10^{-3}\,moles $


Mole of $ Ca(SO _{4}) = \dfrac{136\times 10^{-3}g}{136\,g/mole} = 1\times 10^{3}\,mole $

Total mole of $ Ca = 2\times 10^{-3}\,mole $

mass of $ CaCO _{3} = 2\times 10^{-3}\times 100 = 0.2\,g $

$ \therefore $ ppm (permanent hardness) $ = \dfrac{6.2}{1000}\times 10^{6} = 200\,ppm $

Mole of $ MgCl _{2} = \dfrac{95\times 10^{-3}}{95} = 1\times 10^{-3}\,mole $ 

Mole Mg $ (HCg) _{2} = \dfrac{73\times 10^{-3}}{146} = 5\times 10^{-4}\,mole $
mole Mg $ = 1.5\times 10^{-4}\,mole $

Mole g $ CaCO _{3} $ (In terms of mg) $ = 1.5\times 10^{-3} $

mass $ = 1.5\times 10^{-3} = 0.150\,g $

ppm (temporary hardness) $ = \dfrac{0.150}{100}\times 106 = 150\,ppm $

Hence, the correct option is $\text{C}$