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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. |
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Answer» 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` |
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