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 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 physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

2.56g of sulphur (colloidal sol) in 100 ml solution shows Osmotic pressure of 2.463 atm at ${ 27 }^{ 0 }C$. How many sulphur atoms are associated in colloidal sol ? [Solution constant = 0.0821 atm.${ mol }^{ -1 }{ k }^{ -1 }$]

  1. 2

  2. 4

  3. 5

  4. 8
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the osmotic pressure formula Pi = i * C * R * T, where C = n / V and n = w / M. First find the molar mass of the associated sulfur molecule using the given values: Pi = 2.463 atm, V = 0.1 L, w = 2.56 g, T = 300 K, R = 0.0821. Solving gives the molar mass of the colloidal sulfur molecule as approximately 256 g/mol. Since atomic mass of sulfur is 32, the number of atoms associated is 256 / 32 = 8.

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 $\text{C}$

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

The molecular formula of a commercial resin used for exchanging ions in water softening is $C 8H _7SO _3 $ (Mol. wt. 206). Water would be the maximum uptake of $Ca^{2+} $ ions by the resin when expressed ____ in mole per gram of resin.

  1. $\dfrac{2} {309} $
  2. $\dfrac{1} {412} $
  3. $\dfrac{1} {103} $
  4. $\dfrac{1} {206} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The chemical reaction for softening water can be given as:-


$2C _8H _7(SO _3)Na+Ca^{+2}\longrightarrow [C _8H _7(SO _3)] _2Ca+2Na^{+}$

$\therefore$  $2$ moles of resin$\equiv1$ mole of $Ca^{2+}$

$\therefore$  Mass of resin = $206\times 2=412g$

For $412g$ of resin, $1$ mole of $Ca^{2+}$ is required.

$\therefore$  Maximum uptake of $Ca^{2+}$ ions $= \cfrac {1}{412}$ mole/gram of resin.

Multiple choice chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia
The dissolution of ammonia gas in water does not obey Henry's law. On dissolving, a major portion of ammonia molecules reacts with ${H} _{2}O$ to form ${NH} _{4}OH$ molecules. ${NH} _{4}OH$ again dissociates into ${NH} _{4}^{+}$ and ${OH}^{-}$ ions. In solution therefore, we have ${NH} _{3}$ molecules, ${NH} _{4}OH$ molecules and ${NH} _{4}^{+}$ ions and the following equilibrium exist:

${NH} _{3}(g)$ $ \rightleftharpoons{NH} _{3}(l)+{H} _{2}O{(l)}\rightleftharpoons  {NH} _{4}OH{(aq)}\rightleftharpoons  {NH} _{4}^{+}{(aq)}+{OH}^{-}{(aq)}$

Let, ${c} _{1}$ ${(mol/L)}$ of ${NH} _{3}$ pass in liquid state which on dissolution in water forms ${c} _{2}$ ${(mol/L)}$ of ${NH} _{4}OH$. The solution contains ${c} _{3}$ ${(mol/L)}$ of ${NH} _{4}^{+}$ ions.

Concentration of undissociated ammonium hydroxide is________.
  1. ${c} _{1}+{c} _{2}$
  2. ${c} _{2}-{c} _{3}$
  3. ${c} _{1}+{c} _{3}$
  4. ${c} _{1}-{c} _{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${NH} _{3}(g)$$\rightleftharpoons {NH} _{3}(l)+{H} _{2}O\rightleftharpoons  {NH} _{4}OH\rightleftharpoons  {NH} _{4}^{+}+{OH}^{-}$

$c _1$                                                $c _2$                $c _3$
So, remaining moles of undissociated ammonium hydroxide, $NH _4OH=  c _2-c _3$

Multiple choice chemistry nitrogen and sulfur ammonia-properties and uses ammonia compounds of nitrogen - ammonia

Select the correct options about the solubility of the following species in $ H _2O$ and $ NH _3 $(aq).
$AgF, AgCN, AgCl, Agl, CaF _2, CaCl _2 $ :

  1. $AgF$ and $CaCl _2 $ are soluble in $H _2O$
  2. $AgCN$ and $AgCl$ are soluble in $NH _3$
  3. $AgI$ and $CaF _2$ are soluble in $ H _2O $ and $ NH _3 $ (aq) both

  4. None of these
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

A. $AgF$ and $CaCl _2$ are polar compounds. Hence, they are soluble in water which is a polar solvent. Thus, the option A is correct.

B. $AgCN$ and $AgCl$ form complexes with ammonia as ammonia is a very good ligand. Hence, $AgCN$ and $AgCl$ are soluble in ammonia. Thus, the option B is correct.

C. $CaF _2$ and $AgI$ are insoluble in water. Thus, the option C is incorrect.