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ChemistryMole ConceptMiscellaneous/MixedEasy2 minPYQ_2023
ChemistryEasyassertion reason

25 mLof silver nitrate solution(1M)is added dropwise to25 mLof potassium iodide(1.05 M)solution. The ion(s) present in very small quantity in the solution is/are

Options:

Answer:
D
Solution:

To determine the ion(s) present in very small quantity in the solution, we need to consider the chemical reaction that occurs when silver nitrate reacts with potassium iodide. The reaction between AgNO3 and KI is a double replacement reaction and can be represented as follows:

AgNO3 + KI  AgI + KNO3

Millimoles of AgNO3 = 25

Millimoles of KI = 25 × 1.05=26.25

KI is in excess & AgI forms negatively charged colloid. (Some Ag+remains in solution) lons Ag+& F-are therefore, present in very small quantity.

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Miscellaneous/Mixed
2mℹ️ Source: PYQ_2023

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