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 salts salts and their classification acids, bases and salts chemistry

$0.1\ M\ H _2SO _4$ has the same concentration of $H^+$ ions as $0.1 N HCl$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

0.05 M $\displaystyle H _2SO _4$  has the same concentration of $\displaystyle H^+$ ions as 0.1 M HCl.
 0.05 M $\displaystyle   H _2SO _4$  $\displaystyle  = 2 \times 0.05 = 0.1$ M $\displaystyle H^+$ ions.
 0.1 M HCl $\displaystyle  = 0.1$ M $\displaystyle H^+$ ions.

Multiple choice chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

Common salt obtained from sea-water contains $ 8.775\%$ $NaCl$ by mass. The number of formula units of $NaCl$ present in 25 g of this salt is :

  1. $3.367 \times { 10 }^{ 23 }$ formula units
  2. $2.258 \times { 10 }^{ 22 }$ formula units
  3. $3.176 \times {10 }^{ 23 }$ formula units
  4. $4.73 \times {10 }^{ 25}$ formula units
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First, calculate the mass of NaCl in 25g of salt (25 * 0.08775 = 2.19375g). Then, divide by the molar mass of NaCl (58.44 g/mol) to get moles, and multiply by Avogadro's number (6.022 * 10^23) to find the formula units.

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

Solubility of $MX _{ 2 }$ type electrolytes is $0.5\times 10^{ -4 } mol/L$, Then find out ${ K } _{ sp }$ of electrolytes.

  1. $5\times 10^{ -12 }$
  2. $25\times 10^{ -10 }$
  3. $1\times 10^{ -13 }$
  4. $5\times 10^{ -13 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For an MX2 electrolyte, Ksp = [M^2+][X^-]^2 = (s)(2s)^2 = 4s^3. Given s = 0.5 * 10^-4, Ksp = 4 * (0.5 * 10^-4)^3 = 4 * 0.125 * 10^-12 = 0.5 * 10^-12 = 5 * 10^-13.

Multiple choice chemistry group 17 group 17 - physical properties physical properties of group 17 elements group 17 elements: the halogen family

The solubility of $KCl$ is relatively more in (where D in dielectric constant):

  1. $C _6II _6 (D = 0)$
  2. $(CH _3) _2CO (D = 2)$
  3. $CH _3OH(D= 32)$
  4. $CCl _4 (D = 0)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

KCl is an ionic salt that dissolves well in polar solvents with high dielectric constants. Methanol (CH3OH, D=32) has the highest dielectric constant among the options, making it the best solvent for KCl. Benzene (C6H6), acetone ((CH3)2CO, D=2 is incorrect - should be ~21), and carbon tetrachloride (CCl4) all have very low or zero dielectric constants and are poor solvents for ionic compounds.

Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

$CoCl _{ 3 }.3NH _{ 3 }$ does not form any precipitate with excess ${ AgNO } _{ 3 }$ solution, whereas 1 mole of $CoCl _{ 3 }.5NH _{ 3 }$ gives two moles of $AgCl$ with excess ${ AgNO } _{ 3 }$. The van't Hoff factor for both the compounds respectively are:

  1. 0 and 2

  2. 0 and 3

  3. 1 and 3

  4. 1 and 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To give ppt with ${ AgNO } _{ 3 }$ solution, a substance has to release ${ Cl }^{ - }$ ions.
$CoCl _{ 3 }.3NH _{ 3 }$ gives no ppt because it does not dissociate in solution.
$\therefore $ Number of particles in solution $=1$ i.e the molecule itself $CoCl _{ 3 }.5NH _{ 3 }$ gives 2 moles $AgCl$ ppt because it dissociate as follows
$CoCl _{ 3 }.5NH _{ 3 }\rightarrow \left[ Co\left( { NH } _{ 3 } \right) _{ 5 }Cl \right] ^{ 2+ }+2{ Cl }^{ - }$
It gives 3 ions on dissociation.
$\therefore $ its van't Hoff factor is 3.

Multiple choice chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

On the hydrolysis of $Al{Cl} _{3}$ then the no. of ${Al}^{+3}$ will be:

  1. Is equal to chloride ions
  2. Is equal to $\cfrac{1}{3}$ chloride ions
  3. Three times of ${Cl}^{-}$ ion

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

The hydrolysis reaction will be $AlC{ l } _{ 3 }+{ 3H } _{ 2 }O\rightarrow Al(OH{ ) } _{ 3 }+3HCl$

So the number of $A{ l }^{ +3 }$ will be equal to $\frac { 1 }{ 3 } $ chloride ions.

Multiple choice chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

$AlF _{ 3 }$ is soluble in HF only in presence of KF. its is due to the formation of:

  1. $AlH _{ 3 }$
  2. $K\left[ AlF _{ 3 }H \right]$
  3. ${ K } _{ 3 }\left[ AlF _{ 3 }{ H } _{ 3 } \right]$
  4. ${ K } _{ 3 }\left[ AlF _{ 6 } \right]$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$AlF _3$ is insoluble in the anhydrous $HF$ because the $F$ ions are not available in intermolecular hydrogen bonded $HF$ but it becomes soluble in the  presence of $KF$ due to the formation of soluble complex,

$K _3[AlF _6]$
$AlF _3+3KF\longrightarrow K _3[AlF _6]$

Multiple choice chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

 ${ AlF } _{ 3 }$ is soluble in $HF$ only in presence of $KF$. It is due to the formation of:

  1. ${ AlH } _{ 3 }$
  2. ${ K[AlF } _{ 3 }H]$
  3. ${ { K } _{ 3 }[AlF } _{ 3 }{ H } _{ 3 }]$
  4. ${ { K } _{ 3 }[AlF } _{ 6 }]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$AlF _3$  is insoluble in the anhydrous $HF$ because the $F^-$ ions are not available in intermolecular hydrogen bonded HF, but it becomes soluble in the presence of $KF$ due to the formation of soluble complex, $K _3[AlF _6].$
$AlF _3+3KF⟶K _3[AlF _6]$
Hence option D is correct answer.
Multiple choice chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

Alluminium chloride nexists as dimer, $A{l _2}C{l _6}$ in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives:

  1. ${{\text{[Al(OH}}{{\text{)}} _{\text{6}}}{\text{]}}^{{\text{3 - }}}}{\text{ + 3HCl}}$
  2. ${{\text{[Al(}}{{\text{H}} _2}{\text{O}}{{\text{)}} _{\text{6}}}{\text{]}}^{{\text{3 + }}}}{\text{ + 3HC}}{{\text{l}}^ - }$
  3. $A{l^{3 + }} + 3C{l^ - }$
  4. ${\text{A}}{{\text{l}} _{\text{2}}}{{\text{O}} _{\text{3}}}{\text{ + 6HCl}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When anhydrous aluminum chloride dissolves in water, it undergoes hydration to form the hexaaquaaluminum(III) ion, [Al(H2O)6]3+, and releases chloride ions into the solution.

Multiple choice chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

Aluminium chloride exists as dimer, $Al _2Cl _6$, in solid state as well as in solution of non-polar solvents such as $C _6H _6$.


 When dissolved in water it gives :

  1. $Al _2O _3+6HCl$
  2. $[Al(H _2O) _6]^{3+}+3Cl^-$
  3. $[Al(OH) _6]^{3+}+6Cl^-$
  4. $Al^{3+}+3Cl^-$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Reaction:

$AlCl _3 $ is covalent but in water it becomes ionic due to large hydration energy oh $Al^{3+}$.

$AlCl _3 + 6H _2O \rightarrow [Al(H _2O) _6]^{3+} + 3Cl^-$.


Hence,option B is correct.

Multiple choice chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

Aqueous solution of $AIC{I _3}$ on heating to dryness will give 

  1. $AIC{I _3}$
  2. $A{l _2}{O _3}$
  3. $Al(OH)C{L _2}\,$
  4. $A{L _2}C{l _6}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Heating an aqueous solution of AlCl3 causes hydrolysis where the aluminum ion reacts with water to form aluminum hydroxide, which then dehydrates to aluminum oxide (Al2O3) upon heating to dryness.

Multiple choice chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

$AlCl _3$ achieves stability by forming a dimer. In trivalent state the compound is hydrolysed in water. $AlCl _3$ in acidified aqueous solution forms :

  1. $Al(OH) _3+HCl$
  2. $[Al(H _2O) _6]^{3+}+3Cl^-$
  3. $AlCl _3.2H _2O$
  4. $Al _2O _3+HCl$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Answer:- (B) $[Al({H _2O)} _6]^{+3} +3{Cl}^{-}$

The electron deficiency of Aluminium atom in $Al{Cl} _3$ is compensated by formation of co-ordinate bond between lone pair of Chlorine atom of another $Al{Cl} _3$ molecule and the empty unhybridised p orbital of Aluminium atom, thus forms dimer and in acidified aqueous solution forms $[Al({H _2O)} _6]^{+3} +3{Cl}^{-}$.

$Al{Cl} _3 + 6H _2O \; \longrightarrow \; [Al({H _2O)} _6]^{+3} + 3{Cl}^{-}$
Multiple choice chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

Aluminium chloride exists as dime , $Al _2Cl _6$ in solid state as well as in solution of non-polar solvents like benzene.When dissolved in water, it gives :

  1. $Al^{3+} +2Cl$
  2. $Al _2O _3+6HCl$
  3. $[Al(OH) _6]^3+3HCl$
  4. $[Al(H _2O) _6]^{3+}+3Cl$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Similar to question 430733, the dissolution of AlCl3 in water results in the formation of the octahedral hexaaquaaluminum(III) complex and chloride ions.

Multiple choice chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

The solubility of anhydrous $AlCl _{3}$ and hydrous $AlCl _{3}$ in diethyl ether are $S _{1}$ and $S _{2}$ respectively. Then -

  1. $S _{1}=S _{2}$
  2. $S _{1}> S _{2}$
  3. $S _{1}< S _{2}$
  4. $S _{1}< S _{2}$ but not $S _{1}=S _{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Anhydrous AlCl3 will dissolve the diethyl ether because anhydrous ​AlCl3 is a good lewis acid. It is electron deficient. The lone pairs of diethyl ether will be donated to the anhydrous ​AlCl3 due to which it is soluble. On the other hand, hydrous ​AlCl3 have the water molecules which make it poor lewis acid.


So, $S _{1}> S _{2}$.

Multiple choice chemistry basic analytical techniques filtration and crystallisation other methods of separation separation techniques

In fractional crystallization the proportion of components in the precipitate will depend on their solubility product.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Fractional crystallization is the method of refining substances based on the differences in their solubility. If a mixture of two or more substances in a solution is allowed to crystallise, the least soluble substance will crystallise first. The proportion of the components in the precipitate will depend on their solubility product.