$A$ coordination compound has the formula $MCl_3 \cdot 4H_2O$. The coordination number of the metal is $6$,and it contains no water of hydration. How many $ml$ of $0.1 \, M \, AgNO_3$ are required to precipitate the chloride ions from $200 \, ml$ of a $0.01 \, M$ solution of this compound?

Which of the following statements is correct?

Which of the following change the colour of the aqueous solution of $FeCl_3$?

Given below are two statements:
Statement $I$: $PF_5$ and $BrF_5$ both exhibit $sp^3 d$ hybridisation.
Statement $II$: Both $SF_6$ and $[Co(NH_3)_6]^{3+}$ exhibit $sp^3 d^2$ hybridisation.
In the light of the above statements,choose the correct answer from the options given below:

The reaction of $K_3[Fe(CN)_6]$ with freshly prepared $FeSO_4$ solution produces a dark blue precipitate called Turnbull's blue. The reaction of $K_4[Fe(CN)_6]$ with $FeSO_4$ solution in the complete absence of air produces a white precipitate $X$,which turns blue in air. Mixing the $FeSO_4$ solution with $NaNO_3$,followed by a slow addition of concentrated $H_2SO_4$ through the side of the test tube,produces a brown ring.
Precipitate $X$ is:
$(A)$ $Fe_4[Fe(CN)_6]_3$
$(B)$ $Fe[Fe(CN)_6]$
$(C)$ $K_2Fe[Fe(CN)_6]$
$(D)$ $KFe[Fe(CN)_6]$
Among the following,the brown ring is due to the formation of:
$(A)$ $[Fe(NO)_2(SO_4)_2]^{2-}$
$(B)$ $[Fe(NO)_2(H_2O)_4]^{3+}$
$(C)$ $[Fe(NO)_4(SO_4)_2]$
$(D)$ $[Fe(H_2O)_5(NO)]SO_4$

$A$ mixture $X$ of $0.02 \, mol$ of $[Co(NH_3)_5SO_4]Br$ and $0.02 \, mol$ of $[Co(NH_3)_5Br]SO_4$ was prepared in $2 \, L$ of solution:
$1 \, L$ of mixture $X +$ excess of $AgNO_3 \to Y$
$1 \, L$ of mixture $X +$ excess of $BaCl_2 \to Z$
Number of moles of $Y$ and $Z$ respectively are