500 ml of 0.1 M KCl, 200 ml of 0.01 M $N a N O_{3}$ and 500 ml of 0.1 M $A g N O_{3}$ was mixed. What is the total number of moles of ions in solution?

Q. The volume (in ml) of

0.1

A g N O_{3}

required for complete precipitation of chloride ions present in

30 m l

0.01 M

solution of

[C r (H_{2} O)_{5} C l] C l_{2}

, as silver chloride, is close to:

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In the complex with formula MCl3.4H2O, the coordination number of the metal M is six and there is no molecule of hydration in it. The volume of 0.1 M AgNO3 solution needed to precipitate the free chloride ions in 200 mL of 0.01 M solution of the complex is:

The correct option is B 20 mLAs there is no molecule of hydration in it so compound is M[((H2O)4.Cl2)Cl] which gives a single chloride ion in solution.So no.of moles of silver nitrate = no. of moles of complex 0.1×V=200×0.01 V= 20ml

In the complex with formula $M C l_{3} .4 H_{2} O$ , the coordination number of the metal M is six and there is no molecule of hydration in it. The volume of 0.1 M $A g N O_{3}$ solution needed to precipitate the free chloride ions in 200 mL of 0.01 M solution of the complex is:

The correct option is B 20 mL
As there is no molecule of hydration in it so compound is $M [((H_{2} O)_{4} . C l_{2}) C l]$ which gives a single chloride ion in solution.

So no.of moles of silver nitrate $=$ no. of moles of complex $0.1 \times V = 200 \times 0.01$

V= 20ml