In an ore theonly oxidizable material in Sn^(2+) This is titrated with a dichromate solution containing 2.5 of K_(2)Cr_(2)O_(7) in 0.50 litre A 0.40 g sample of the ore required 10.0 cm^(3) of titrant to reach equivalenc point calculate the prcentage of tin in the re (k=39.1, Cr = 52 Sn =11.87)

In an ore theonly oxidizable material in Sn^(2+) This is titrated with

1.	In an ore theonly oxidizable material in Sn^(2+) This is titrated with a dichromate solution containing 2.5 of K_(2)Cr_(2)O_(7) in 0.50 litre A 0.40 g sample of the ore required 10.0 cm^(3) of titrant to reach equivalenc point calculate the prcentage of tin in the re (k=39.1, Cr = 52 Sn =11.87)
Answer» Solution :Step 1 To write balanced chemical EQUATION f the redox reaction `Sn^(2+)rarr Sn^(4+)+2 E^(-)[xx3` `Cr_(2)O_(7)^(2-)+14 H^(+)+6 e^(-) rarr 2Cr^(3+)+7 H_(2)O` `Cr_(2)O_(7)^(2-)+3 Sn^(2+)+14 H^(+) rarr 2Cr^(3+)+3 Sn^(4+)+7 H_(2)O` Now 0.5 L of the solution cotains `K_(2)Cr_(2)O_(7)=2.5 g` `therefore` 10 mL of solution will contain `K_(2)Cr_(2)O_(7)=(2.5)/(500)xx10 g=(2.5)/(500xx(210)/(294)` mole From the balanced redox reaction 1 mole of `K_(2)Cr_(2)O_(7)` oxidises `Sn^(2+)=3` moles `therefore (2.5)/(500)XX(10)/(294)` mole of `K_(2)Cr_(2)O_(7)` will oxidise `Sn^(2+)=3xx(2.5)/(500)xx(10)/(294) mol =3x(2.5)/(500)xx(10)/(294)xx118.7 g` Step 2 To determine the % age of `Sn^(2+)` in the ore Now 0.40 g of the or contain `Sn^(2+) =0.06 g` `therefore %"of"Sn^(2+) "in theare" = 0.06/0.40xx100=15%`