Q. A solution containing 2.675g of CoCl3. 6NH3 (molar mass = 267.5 g/mol) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.78g of AgCl(molar mass = 143.5 g/mol). The formula of the complex is? (At mass of Ag = 108u)
I didn't understand the FIITJEE solution, so if someone could please explain the approach to this question, I would be grateful.
1i dint get the ans too...this is how i proceeded..
the number of ionisable Cl- ions per molecule of the complex is important as they are rthose ions which form the ppt with AgNO3.
Cl- + AgNO3 → AgCl + NO3-
35.5 143.5
143.5 g of AgCl is given by 35.5 g of Cl- ions
4.78 g of AgCl is given by 1.18 g of Cl- ions
let the no of ionisable Cl- ions per mole of complex be x.
no of moles of complex=0.01
wt of Cl- in 1 mole=35.5x
wt of Cl- in 0.01 moles=0.355x
0.355x=1.18
x=3.3 not getting the answer[2]
1what was the answer given pritish?
someone plz tell me my mistake,or provide the solution for this sum..
1this was the easiest question of the question paper.
u will see that only the 2nd option had 6 molecules of NH3.so rest have to be wrong.
hope i had checked that earlier[2][2][2][2][2]
but just to verify the correctness of the answer,dont u think my method is correct..and i dont think 3.3 can be approximated as 3 can it?
13We've simply got to equate the wt. of chlorine we get frm both.
n x 2.675267.5 x 35.5 = 35.5143.5 x (4.78)
(where n = moles of Cl- ions we get frm ionisation of 1mole complx.)
which gives "n" ≈3
13Oh sry...was in such a hurry ki didnt see Arka had already posted the solution :O
Thts right arka, it is exactly wht i've done. [1]
