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

According to conductance measurement, how many ions are present in a dilute solution of CoCl3.4NH3?

  1. 2

  2. 3

  3. 5

  4. 6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

CoCl3.4NH3 dissociates in water to give [Co(NH3)4Cl2]+ and Cl- ions, totaling 2 ions per formula unit. Conductance measurements directly count the number of ions produced, not the coordination number. The compound exists as [Co(NH3)4Cl2]Cl, where the complex cation and one chloride ion separate in solution.

Multiple choice

The volume (in mL) of 0.2 M AgNO3 required for complete precipitation of chloride ions present in 50 mL of 0.02 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is __________.

  1. 1 mL

  2. 2 mL

  3. 5 mL

  4. 10 mL

  5. 20 mL
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

[Cr(H2O)5Cl]Cl2 + 2AgNO3 → 2AgCl + [Cr(H2O)5Cl](NO32 Number of ionisable chloride ions in complex [Cr(H2O)5Cl]Cl2 = 2. Millimoles = Molarity x Volume (mL) x 2= 0.02 x 50 x 2 = 2.0 Therefore, required Ag + ions = 2.0 millimoles Millimoles = Molarity x V (mL) 2.0 = 0.2 x V V = 10 mL

Multiple choice

What is the correct order of ion mobility of the following ions in an aqueous solution?

  1. Cation: H3O+ > Li+ > Na+, Anion: OH- > Br- > CH3COO-

  2. Cation: H3O+ > Na+ > Li+, Anion: OH- > Br- > CH3COO-

  3. Cation: Na+ > Li+ > H3O+, Anion: OH- > Br- > CH3COO-

  4. Cation: H3O+ > Li+ > Na+, Anion: CH3COO- > Br- > OH-

  5. Cation: H3O+ > Na+ > Li+, Anion: Br- > OH-> CH3COO-
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is the correct order of movement of anions and cations in an aqueous solution.

Multiple choice

Melting point of NaCl is ______.

  1. 800°C

  2. 825°C

  3. 850°C

  4. None of these

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sodium chloride (NaCl) is a common ionic compound with a high melting point, specifically around 801 degrees Celsius.

Multiple choice

In which type of cell zinc and copper plates are dipped in separate solution?

  1. Voltaic cell

  2. Dry cell

  3. Plant cell

  4. Lachlan he cell
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A galvanic cell, or voltaic cell is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within the cell. It generally consists of two different metals connected by a salt bridge ( copper and zinc).

Multiple choice

The solubility of sodium bicarbonate in water is 9.6 g/100 g at 20°C and 16.4 g/100 g at 60°C. If a saturated solution of sodium bicarbonate at 60°C is cooled to 20°C, the percentage of the dissolved salt crystallized out will be

  1. 20.5

  2. 25.4

  3. 41.5

  4. 45.2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The percentage of salt crystallized out is calculated based on the difference in solubility between 60 degrees Celsius and 20 degrees Celsius per unit of solvent. Starting with 16.4 g of solute per 100 g of water at the higher temperature, dropping to 9.6 g means 16.4 minus 9.6 equals 6.8 g precipitates out of the 16.4 g initially dissolved. Calculating 6.8 divided by 16.4 and multiplying by 100 gives approximately 41.5 percent.

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 $MgBa(CO _3) _2$.