InterviewSolution
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InterviewSolution
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In an ore, the only oxidizable material is `Sn^(2+)`. This ore is titrated with a dichromate solution containing `2.5g` of `K_(2)Cr_(2)O_(7)` in `0.5 "litre"`. A `0.40 g` sample of the ore required `10.0 cm^(3)` of titrant to reach equivalence point. Calculate the percentage of tin in ore. |
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Answer» Redox changes are: `Sn^(2+)rarrSn^(4+)+2e` `6e+Cr_(2)^(6+)rarr2Cr^(3+)` Since `Sn^(2+)` is oxidised by `K_(2)Cr_(2)O_(7)`. `therefore` Meq.of `Sn^(2+)="Meq.of" K_(2)Cr_(2)O_(7)` used for tin `=NxxV_("in " mL) ( because N=(2.5)/((294.2)/(6)xx0.5))` `= (2.5)/((294.2)/(6)xx0.50)xx10=1.0197` `therefore (w_(Sn^(2+)))/(118//2)xx1000=1.0197` `therefore w_(Sn^(2+))=0.06g` `therefore % Sn=(0.06)/(0.4)xx100=15%` |
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