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 the s-block elements (alkali and alkaline earth metals) uses of alkali and alkaline earth metals uses of s block elements group 1 elements: alkali metals

Potash alum is represented by the formula :

  1. $K _2SO _4\cdot Al _2(SO _4) _3\cdot 24H _2O$
  2. $K\cdot Al(SO _4) _2\cdot 12 H _2O$

  3. Both (a) and (b)

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

Potassium alum or potassium aluminium sulphate is a chemical compound with a chemical formula $KAl(SO _4) _2$ and it is commonly found in its dodecahydrate form as $KAl(SO _4) _2.12H _2O$. 


Alum is the common name for this chemical compound.

Multiple choice limestone uses of group 2 compounds group 2 industrial inorganic chemistry chemistry

Based on the following analytical data, answer the given question.
A mineral, which can be represented by the formula $Mg _xBa _y(CO _3) _2$, was analyzed as described below:

A sample of the mineral was dissolved in excess hydrochloric acid and the solution made up to $100 cm^3$ with water. During the process, $48 cm^3$ of carbon dioxide, measured at $25^{\circ}C$ and 1-atmosphere pressure, were evolved.
A $25.0 cm^3$ portion of the resulting solution required $25.0 cm^3$ of EDTA solution of concentration $0.02 \ mol / dm^3$ to reach an end-point. A further $25.0 cm^3$ portion gave a precipitate of barium sulphate of mass 0.058 g on treatment with excess dilute sulphuric acid. You may assume that group-2 metal ions form 1:1 complexes with EDTA. Molar volume of any gas at $25^{\circ}C$ and 1 atmosphere pressure $ = 24 dm^3$).

The formula of the mineral is: 

  1. $MgBa(CO _3) _4$
  2. $MgBa(CO _3) _2$
  3. $MgBa(CO _3) _3$
  4. $Mg _2Ba(CO _3) _4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The formula of the mineral is $MgBa(CO _3) _2$.

$48 cm^3$ of carbon dioxide corresponds to 0.002 moles.
Thus, $MgBa(CO _3) _2$ reacts with excess HCl to give 0.002 moles of carbon dioxide.

So $100.0cm^3$ solution will give 0.002 moles of carbon dioxide.
$25.0cm^3$ solution will give 0.0005 moles of carbon dioxide.

A $25.0cm^3$ portion of the resulting solution required $25.0cm^3$ of EDTA solution of concentration $0.02mol.dm^{-3}$ to reach an end-point.

Thus $25.0cm^3$ portion of the resulting solution contains $0.02 \times \frac {25.0}{1000}=0.0005 mol$ of metal ions.

A further $25.0cm^3$ portion gave a precipitate of barium sulphate of mass 0.058 g on treatment with excess dilute sulphuric acid.

0.058 g of barium sulphate corresponds to $\frac {0.058 g}{233.43  g/mol}=0.00025\  mol$.

$\therefore $  Moles of $Ba^{2+}$ ion $=$ Moles of $Mg^{2+}$ ion $=0.00025\ mol$  

So, formula of the compound is $MgBa(CO _3) _2$.

Multiple choice chemistry the s-block elements (alkali and alkaline earth metals) biological importance of magnesium and calcium biological importance of elements biological importance of s block elements biological importance of sodium and potassium

For a given sample of water containing the following impurities
$Mg{ \left( H{ CO } _{ 3 } \right)  } _{ 2 }=73mg/L;\quad Ca{ \left( H{ CO } _{ 3 } \right)  } _{ 2 }=162mg/L;\quad Ca{ SO } _{ 4 }=136mg/L$
$Mg{ Cl } _{ 2 }=95mg/L;\quad Ca{ Cl } _{ 2 }=111mg/L;\quad NaCl=100mg/L$
Then the total hardness (temporary and permanent) of above water sample is

  1. $300ppm$
  2. $350ppm$
  3. $450ppm$
  4. $500ppm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total hardness is calculated based on the concentration of Ca2+ and Mg2+ salts expressed as CaCO3 equivalents. Using molar masses (Mg(HCO3)2=146, Ca(HCO3)2=162, CaSO4=136, MgCl2=95, CaCl2=111), the calculation yields 300 ppm.

Multiple choice reactions of acids and bases properties of acids and bases acids, bases and salts acids and bases chemistry

The chloride salt of a certain weak monoacidic organic base is hydrolysed to an extent of $3$% in its $0.1M$ solution at ${25}^{o}C$. Given that the ionic product of water is ${10}^{-14}$ at this temperature, what is the dissociation constant of the base?

  1. $\approx 1\times { 10 }^{ -10 }$
  2. $\approx 1\times { 10 }^{ -9 }$
  3. $\approx 3.33\times { 10 }^{ -9 }$
  4. $\approx 3.33\times { 10 }^{ -10 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$h = 0.03$, $C = 0.1 M$ , $K _{w} = 10^{-14}$       

We know, $K _{a} = Ch^{2}   $
$\cfrac{K _{w}}{K _{b}} = Ch^{2}    \Rightarrow \cfrac{10^{-14}}{K _{b}} = (0.1)(0.03)^{2} \Rightarrow  \cfrac{10^{-9}}{9} \approx 1 \times 10^{-10}$

Multiple choice reactions of acids and bases properties of acids and bases acids, bases and salts acids and bases chemistry

A mixture containing one mole of $BaCl _2$ and two moles of $H _2SO _4$ will be neutralized by:

  1. $1$ mole KOH
  2. $4$ mole KOH
  3. $2$ mole KOH
  4. $1$ mole Ca(OH)$ _2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

BaCl2 + H2SO4 -> BaSO4 + 2HCl. One mole of BaCl2 reacts with one mole of H2SO4 to produce 2 moles of HCl. The remaining 1 mole of H2SO4 is also acidic. Total acid to neutralize is 2 moles of HCl + 1 mole of H2SO4 (which has 2 H+), totaling 4 moles of H+. Therefore, 4 moles of KOH are required.

Multiple choice introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

A mixture contains $NaCl$ and unknown chloride $MCl$. When $1\ g$ of this mixture is dissolved in water and excess of $AgNO _{3}$ Solution is added to it, $2.567\ g$ of white precipitate is obtained. In another experiment, $1\ g$ of the same original mixture is heated to $300^{o}C$. Some vapour come out which are absorbed in acidified $AgNO _{3}$ solution by which $1.341\ g$ of white precipitate is formed. The molecular mass of unknown chloride is

  1. $53.4$
  2. $58.5$
  3. $44.5$
  4. $74.4$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Multiple choice chemistry p- block elements-ii compounds of phosphorus - pcl3 phosphorus halides compounds of phosphorus compounds of phosphorus- pcl5

Aqueous solution of PBr$ _{3}$ conducts electricity due to the presence of:

  1. $\mathrm{HOBr}$
  2. $\mathrm{HBr}$
  3. $\mathrm{H} _{3}\mathrm{P}\mathrm{O} _{4}$
  4. $\mathrm{H} _{2}\mathrm{O}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$PBr _{3}+3H _{2}O\ \to \ H _{3}PO _{4}+HBr$
HBr is covalent $H _{3}PO _{4}$ is ionic. Solution conducts electricity due to ions.
So aqueous solution of $PBr _{3}$ conducts electricity due to presence of $H _{3}PO _{4}$.

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

If the percentage yield of the $1 st$ step is $80\% $ and that of the $2nd $ step is $75\% $, then what is the expected overall percentage yield for producing $CaO _3$ from $CaCl _{2} $?

  1. $50\%$
  2. $70\%$
  3. $55\%$
  4. $60\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
${ CaCl } _{ 2 }\xrightarrow [  ]{ { 1 }^{ st }step } \times \xrightarrow [  ]{ { 2 }^{ nd }step } { CaCO } _{ 3 }$
Let $100$ unit of $CaCl _2$ taken
amount of $\times $ produced $=80\ unit$
amount of $CaCO _3$ produced $=\dfrac {80\times 75}{100}=60$
Hence $\%$ yield of $CaCO _3$ by $CaCl _2=60\%$
Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

To a $10$ml $1M$ aqueous solution of $Br _2$,excess of NaOH is added so that all $Br _2$ is disproportional to $Br^-$ and $BrO _3^-$, the resulting solution is freed from $Br^-$,by extraction and excess of $OH^-$ neutralised by acidifying the solution. The resulting solution is sufficient the react with $1.5$gm of impure $CaC _2I _4$ $(M=128gm /mol)$ sample. The purity by mass of Oxalate sample is the relevant reaction s are $Br _2(aq.)+OH^- \rightarrow (aq.)+BrO _3^-$
$Bro _3^-+C _2O _4^{2-}\rightarrow Br^-+CO _2$

  1. $85.3\%$
  2. $12.5\%$
  3. $90\%$
  4. $50\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Br2 + 6OH- -> 5Br- + BrO3- + 3H2O. 10 ml 1M Br2 = 0.01 mol. This produces 0.01/6 = 0.00166 mol BrO3-. Reaction: 2BrO3- + 3C2O4(2-) -> 2Br- + 6CO2. 0.00166 mol BrO3- reacts with 0.0025 mol C2O4(2-). 0.0025 mol * 128 g/mol = 0.32 g. Purity = (0.32 / 1.5) * 100 = 21.3%. The provided answer 85.3% seems to rely on different stoichiometry.

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

$0.5\ g$ of impure ammonium chloride was heated with caustic soda solution to evolve ammonia gas, the gas is absorbed in $150\ mL$ of $N/5\ H _{2}SO _{4}$ solution. Excess sulphuric acid required $20\ mL$ of $1\ N\ NaOH$ for complete neutralization. The percentage of $NH _{3}$ in the ammonium chloride is:

  1. $68$%
  2. $34$%
  3. $48$%
  4. $17$%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$(NH _3)Cl+NaOH\rightarrow +NH _3\uparrow +H _2O$

excess $H _2SO _4$ reg. $20ml$ of $1N\,\,NaOH$
$\Rightarrow $ moles of $H _2SO _4=(20\times 1)milimoles$
                                 $=20m\,mol$
added amount of $H _2SO _4=\dfrac{N}{5}\times 150ml=30mmol$
amount of $h _2so _4$ reacted with $nh _3=0.01MOLE$
$\Rightarrow 0.01\, mole$ of $NH _3$ present in $(NH _4)Cl$
So, $\%purity=\dfrac{(0.01)\times 17}{0.5}\times 100=34\%$