To find formulae of a compound if we titrate `NH_(3)` in the compound with standardized acid.
`Cr(NH_(3))_(x)Cl_(3)(aq)+xHCl(aq)rarrxNH_(4)^(+)(aq)+Cr^(3+)(aq)+(x+3)Cl^(-)(aq)`
Assume that 20ml of 1.5 MHCl is used to titrate 1.3025 gm of `Cr(NH_(3))_(x)Cl_(3)` what is value of x.
Express your answer as the nearest integer value.

Explanation

20ml 1.5MHCl
`rArr 20xx1.5xx10^(-3)=0.03` mol HCl
moles of `Cr(NH_(3))_(x)Cl_(3)=("moles of HCl")/(x)`
`=(0.03)/(x)`
so `(0.03)/(x)=(1.3025)/(55+106.5+17x)`
`0.51x +4.845-1.3025x`
`4.845=(1.3025-0.51)x`
`4.845=0.7925x`
`x=(4.845)/(0.7925)=6.11`