Home
Class 12
CHEMISTRY
The variation in Lambdam with concentrat...

The variation in `Lambda_m` with concentration for a strong electrolyte can be represented by the equation, `Lambda_m=Lambda_m^@ -AC^(1//2)` The value of constant A for a given solvent and temperature depends upon the type of electrolyte i.e., cations and anions produced on dissociation of electrolyte in the solution .
NaCl, `MgCl_2` and `CaSO_4` are known as

A

1-1,2-1, and 2-2 type electrolytes respectively

B

strong, weak and strong electrolytes respectively

C

electrolytes with different value of A

D

electrolytes with same molar conductivity .

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the classification of NaCl, MgCl₂, and CaSO₄ based on their dissociation into cations and anions, we will analyze each electrolyte step by step. ### Step-by-Step Solution: 1. **Understanding Molar Conductivity (Λm)**: - Molar conductivity (Λm) is a measure of the ability of an electrolyte to conduct electricity when dissolved in a solvent. It depends on the concentration of ions in the solution. 2. **Equation for Molar Conductivity**: - The variation of molar conductivity with concentration for a strong electrolyte is given by the equation: \[ \Lambda_m = \Lambda_m^0 - A C^{1/2} \] - Here, Λm^0 is the molar conductivity at infinite dilution, A is a constant, and C is the concentration of the electrolyte. 3. **Identifying the Electrolytes**: - The electrolytes given are: - NaCl (Sodium Chloride) - MgCl₂ (Magnesium Chloride) - CaSO₄ (Calcium Sulfate) 4. **Dissociation of Electrolytes**: - **NaCl**: Dissociates into Na⁺ and Cl⁻. - **Type**: 1:1 electrolyte (1 cation and 1 anion). - **MgCl₂**: Dissociates into Mg²⁺ and 2 Cl⁻. - **Type**: 1:2 electrolyte (1 cation and 2 anions). - **CaSO₄**: Dissociates into Ca²⁺ and SO₄²⁻. - **Type**: 1:1 electrolyte (1 cation and 1 anion). 5. **Classification of Electrolytes**: - Based on their dissociation: - NaCl is a **1:1 electrolyte**. - MgCl₂ is a **1:2 electrolyte**. - CaSO₄ is a **1:1 electrolyte**. 6. **Conclusion**: - Therefore, NaCl, MgCl₂, and CaSO₄ can be classified as: - 1:1 (NaCl), 1:2 (MgCl₂), and 1:1 (CaSO₄) electrolytes respectively. ### Final Answer: NaCl is a 1:1 electrolyte, MgCl₂ is a 1:2 electrolyte, and CaSO₄ is a 1:1 electrolyte. ---
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    NCERT FINGERTIPS ENGLISH|Exercise HOTS|10 Videos
  • ELECTROCHEMISTRY

    NCERT FINGERTIPS ENGLISH|Exercise EXEMPLAR PROBLEMS|17 Videos
  • COORDINATION COMPOUNDS

    NCERT FINGERTIPS ENGLISH|Exercise Assertion And Reason|15 Videos
  • GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

    NCERT FINGERTIPS ENGLISH|Exercise Assertion And Reason|15 Videos

Similar Questions

Explore conceptually related problems

The variation in Lambda_m with concentration for a strong electrolyte can be represented by the equation, Lambda_m=Lambda_m^@ -AC^(1//2) The value of constant A for a given solvent and temperature depends upon the type of electrolyte i.e., cations and anions produced on dissociation of electrolyte in the solution . Which of the following statements is correct regarding variations of molar conductivity with concentration.

The molar conductivity at infinite dilution of a strong electrolyte can be obtained by extrapolating the curve lambda_(m) versus sqrt(c ) but same cannot be done for a weak electrolyte. Explain.

Resistance of 0.2 M solution of an electrolyte is 50 ohm . The specific conductance of the solution is 1.4 S m^(-1) . The resistance of 0.5 M solution of the same electrolyte is 280 Omega . The molar conductivity of 0.5 M solution of the electrolyte in S m^(2) mol^(-1) id

Molal depression constant for a solvent is 4.0 kg mol^(-1) . The depression in the freezing point of the solvent for 0.03 mol kg^(-1) solution of K_(2)SO_(4) is : (Assume complete dissociation of the electrolyte)

Solutions of two electrolytes A and B are diluted. The Lambda_(m) of 'B' increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer.

The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation lambda_m^( C)=lambda_m^(oo)-bsqrtC where lambda_(m)^( C) =molar specific conductance lambda_(m)^(oo) =molar specific conductance at infinite dilution C=molar concentration {:("Molar Concentration of NaCl","Molar Conductance" "In" "ohm"^(-1)cm^(2)"mole"^(-1)),(4xx10^(-4),107),(9xx10^(-4),97),(16xx10^(-4),87):} When a certain conductivity cell (C) was filled with 25xx10^(-4) (M) NaCl solution.The resistance of the cell was found to be 1000 ohm.At Infinite dilution, conductance of Cl^(-) and SO_4^(-2) are 80 ohm^(-1) cm^(2) "mole"^(-1) and 160 ohm^(-1) cm^2 "mole"^(-1) respectively. If the cell ( C) is filled with 5xx10^(-3)(N)Na_(2)SO_(4) the obserbed resistance was 400 ohm.What is the molar conductance of Na_(2)SO_(4) .

A solution which remains in equilibrium with undissolved solute , in contact , is said to be saturated . The concentration of a saturated solution at a given temperature is a called solubility . The product of concentration of ions in a saturated solution of an electrolyte at a given temperature is called solubility product (K_(sp)) . For the electrolyte A_(x),B_(y) with solubility S. The solubility product (K_(sp)) is given as K_(sp) = x^(x) xx y^(y) xx S^(x-y) . While calculating the solubility of a sparingly . soluable salt in the presence of some strong electrolyte containing a common ion , the common ion concentration is practically equal to that of strong electrolyte containing a common ion . the common ion soncentration is practically equal to that of strong electrolyte . If in a solution , the ionic product of an electroylte exceeds its K_(sp) value at a particular temperature , then precipitation occurs . If two or more electrolyte are presentt in the solution , then by the addition of some suitable reagent , precipitation generally occurs in increasing order of their k_(sp) values . Solubility of some sparingly soluable salts , is sometimes enhanced through complexation . While we are calculating the solubility of some sparingly or pH of an electrolyte , the nature of cation of anion should be checked carefully whether there ion (s) are capable of undergoing hydrolysis or not . If either or both of the ions are capable of undergoing hydrolysis , it should be taken into account in calculating the solubility . While calculating the pH of an amphiprotic species , it should be checked whether or not cation can undergo hydrolysis . Total a_(H^(-)) = sqrt(K_(a_(1)xxK_(a_(2)))) (if cation do not undergo hydrolysis ) a_(H^(+)) = sqrt(K_(a_(1))((K_(w))/(K_(b)) - K_(a_(2)))) (if cation also undergoes hydrolysis ) where symbols have usual meaning . Solubility of solids into liquids is a function of temperature alone but solubility of gases into liquids is a function of temperature as well as pressure . The effect of pressure on solubility of gases into liquids is governed by Henry's law . The solubility of BaSO_(4) in 0.1 M BaCl_(2) solution is (K_(sp) " of " BaSO_(4) = 1.5 xx 10^(-9))

A solution which remains in equilibrium with undissolved solute , in contact , is said to be saturated . The concentration of a saturated solution at a given temperature is a called solubility . The product of concentration of ions in a saturated solution of an electrolyte at a given temperature is called solubility product (K_(sp)) . For the electrolyte A_(x),B_(y) with solubility S. The solubility product (K_(sp)) is given as K_(sp) = x^(x) xx y^(y) xx S^(x-y) . While calculating the solubility of a sparingly . soluable salt in the presence of some strong electrolyte containing a common ion , the common ion concentration is practically equal to that of strong electrolyte containing a common ion . the common ion soncentration is practically equal to that of strong electrolyte . If in a solution , the ionic product of an electroylte exceeds its K_(sp) value at a particular temperature , then precipitation occurs . If two or more electrolyte are presentt in the solution , then by the addition of some suitable reagent , precipitation generally occurs in increasing order of their k_(sp) values . Solubility of some sparingly soluable salts , is sometimes enhanced through complexation . While we are calculating the solubility of some sparingly or pH of an electrolyte , the nature of cation of anion should be checked carefully whether there ion (s) are capable of undergoing hydrolysis or not . If either or both of the ions are capable of undergoing hydrolysis , it should be taken into account in calculating the solubility . While calculating the pH of an amphiprotic species , it should be checked whether or not cation can undergo hydrolysis . Total a_(H^(-)) = sqrt(K_(a_(1)xxK_(a_(2)))) (if cation do not undergo hydrolysis ) a_(H^(+)) = sqrt(K_(a_(1))((K_(w))/(K_(b)) - K_(a_(2)))) (if cation also undergoes hydrolysis ) where symbols have usual meaning . Solubility of solids into liquids is a function of temperature alone but solubility of gases into liquids is a function of temperature as well as pressure . The effect of pressure on solubility of gases into liquids is governed by Henry's law . The solubility of PbSO_(4) in water is 0.0303 g/l at 25^(@)C , its solubility product at that temperature is

Resistance of 0.2 M solution of an electrolyte is 50 Omega . The specific conductance of the solution is 1.3 S m^(-1) . If resistance of the 0.4 M solution of the same electrolyte is 260 Omega , its molar conductivity is .

NCERT FINGERTIPS ENGLISH-ELECTROCHEMISTRY-Assertion And Reason
  1. The variation in Lambdam with concentration for a strong electrolyte c...

    Text Solution

    |

  2. Assertion:Electrolytic cell uses electrical energy to carry non-sponta...

    Text Solution

    |

  3. Assertion :EMF of the cell is the potential difference between the ele...

    Text Solution

    |

  4. Assertion:A standard hydrogen electrode is also called reversible elec...

    Text Solution

    |

  5. Assertion:Cu^(2+) ions get reduced more easily than H^+ ions. Reaso...

    Text Solution

    |

  6. Assertion:Lithium has the lowest electrode potential. Reason: Lithiu...

    Text Solution

    |

  7. Assertion : Current stops flowing when E(cell)=0. Reason : ...

    Text Solution

    |

  8. Assertion : To obtain maximum work from a galvanic cell charge has to...

    Text Solution

    |

  9. Assertion:The electrical resistance of any object decrease with increa...

    Text Solution

    |

  10. Assertion: The conductivity of electrolytic soutions increase with inc...

    Text Solution

    |

  11. Assertion:Molar conductivity increases with decrease in concentration....

    Text Solution

    |

  12. Assertion:Kohlrausch law helps to find the molar conductivity of weak ...

    Text Solution

    |

  13. Assertion:When a copper wire is dipped in silver nitrate solution, the...

    Text Solution

    |

  14. Assertion : In electrolysis, the quantity of electricity needed for de...

    Text Solution

    |

  15. Assertion:In electrolysis of aqueous NaCl the product obtained is H2 g...

    Text Solution

    |

  16. Assertion:In mercury cell, the cell potential is approximately 1.35 V ...

    Text Solution

    |