Home
Class 12
CHEMISTRY
When 0.1 mol of CoCl(3)(NH(3))(5) is tre...

When 0.1 mol of `CoCl_(3)(NH_(3))_(5)` is treated with excess of `AgNO_(3), 0.2` mol of Agcl is obtained. The conductivity of solution will correspond to

A

1:3 electroylte

B

1:2 electroyle

C

1:1 electrolyte

D

3:1 electrolyte

Text Solution

Verified by Experts

0.2 mol of Agcl is obtained when 0.1 mol of `CoCl_(3)(NH_(3))_(5)` is treatd with excess of `AgNO_(3)` which shows that one molecule of the complex gives two `Cl^(-)` ions in solution. Thus the formula of the complex is `[Co(NH_(3))_(5)Cl]Cl_(2)` i.e. 1:2 electrolyge.
Promotional Banner

Topper's Solved these Questions

  • COORDINATION COMPOUNDS

    MTG-WBJEE|Exercise WBJEE/WORKOUT (Category 2: Single Option Correct Type)|13 Videos

  • COORDINATION COMPOUNDS

    MTG-WBJEE|Exercise WBJEE/WORKOUT (Category 3: One or More than one Option Correct Type)|7 Videos

  • CHEMISTRY OF NON-METALLIC ELEMENTS AND THEIR COMPOUNDS

    MTG-WBJEE|Exercise WB JEE PREVIOUS YEARS QUESTIONS(CATEGORY 3 : One or More Option Correct Type)|1 Videos

  • ENVIRONMENTAL CHEMISTRY

    MTG-WBJEE|Exercise WB JEE Previous Years Questions ( CATEGORY 1 : Single Option Correct Type )|2 Videos

Similar Questions

Explore conceptually related problems

When 0.1 mol CoCl_(3)(NH_(3))_(5) is treated with excess of AgNO_(3) , 0.2 mole of AgCl are obtained. The conductivity of solution will correspond to

When 1 mol CrCl_(3).6H_(2)O is treated with excess of AgNO_(3) , 3 mol of AgCl are obtained. The formula of the coplex is

When one mole CrCl_3 .6H_2O is treated with excess of AgNo_3 , 3 mole of AgCl is obtained. The formula of complex is