The volume of 0.1M AgNO3 required for complete precipitation of chloride ions present in 30 ml of 0.01M solution of [Cr(H2O)5Cl]Cl2 , as silver chloride is ?

Question

Answer

The chemical reaction between them is...[Cr(H2O)5Cl]Cl2 + {2}AgNO3 ——>  {2}AgCl(↓)  +  [Cr(H2O)5Cl]²+  +  {2}NO3-moles of coordinate compound given = molarity × volume(in litre)mole = 0.01 × (30/1000) = (3/10000)for one mole coordinate compound.. we need 2 mole AgNO3.....i.e.,1mole complex ===> 2 mole AgNO3(3/10000) mole ==> (2×3/10000) mole AgNO3molarity = moles/volume (in litre)volume(in litre) = mole / molarityV= (6/10000) ÷ 0.1 = (0.6) Litrehence volume required = 0.6litre = 6ml

Answer

Its 6 ml.as you know the chloride ions inside the coordination sphere cannot be precipitated.so 1 mole of complex contains 2 ppt chloride ions.apply n1v1=2n2v2 where n2v2 is for complex.

Answer

**Calculation of moles of chloride ions**

To determine the volume of 0.1M AgNO3 required for the complete precipitation of chloride ions, we need to first calculate the number of moles of chloride ions present in the solution.

Given:
- Volume of the solution = 30 ml = 0.03 L
- Concentration of [Cr(H2O)5Cl]Cl2 = 0.01M

[Cr(H2O)5Cl]Cl2 is a complex compound containing one chloride ion per formula unit. Therefore, the concentration of chloride ions is also 0.01M.

Number of moles of chloride ions = concentration × volume
= 0.01M × 0.03 L
= 0.0003 moles

**Stoichiometry of the reaction**

The reaction between silver nitrate (AgNO3) and chloride ions (Cl-) forms silver chloride (AgCl) precipitate.

The balanced equation for the reaction is:
AgNO3 + Cl- → AgCl + NO3-

From the stoichiometry of the reaction, we can see that 1 mole of AgNO3 reacts with 1 mole of Cl- to form 1 mole of AgCl.

Therefore, the number of moles of AgNO3 required for the complete precipitation of chloride ions is also 0.0003 moles.

**Calculation of volume of 0.1M AgNO3**

The concentration of AgNO3 is given as 0.1M, which means that there are 0.1 moles of AgNO3 in 1 liter of solution.

Number of moles of AgNO3 required = 0.0003 moles

Volume of AgNO3 required = moles/concentration
= 0.0003 moles / 0.1M
= 0.003 L = 3 ml

Therefore, the volume of 0.1M AgNO3 required for the complete precipitation of chloride ions present in 30 ml of 0.01M [Cr(H2O)5Cl]Cl2 solution is 3 ml.

This volume of AgNO3 is necessary to react with all the chloride ions in the solution and form silver chloride precipitate.

Note: The silver chloride precipitate can be separated from the solution by filtration, and its mass can be determined to calculate the percentage yield of the reaction.

Answer

6

Answer

I guess its 6.

Answer

Its..6

Answer

Haa ..may be it is 6.. .. i too got the same

Answer

6 mL

Answer

ARRE SO EASY, IT US DEFINITELY GOING TO BE 6 ML.

Answer

6ml
30×0.01= 0.3mmol
2×0.3 =0.6 mmol are req
0.6÷0.1 = 6 ml

Answer

The answer is 6 ml

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