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A mixture containing As_(2)O_(3) and As_(2)O_(3) requried 20 " mL of " 0.05 N I_(2) for titration. The resulting solution is then acidified and excess KI was added. The liberated I_(2) required 1.24 g hypo (Na_(2)S_(2)O_(3).H_(2)O) for complete reaction. Calculate the mass of the mixture. The reactions are As_(2)O_(3)+2I_(2)+2H_(2)OtoAs_(2)O_(3)+4H^(o+)+4I^(ɵ) As_(2)O_(5)+4H^(o+)+4I^(ɵ)toAs_(2)O_(3)+2I_(2)+2H_(2)O

Answer» <html><body><p></p>Solution :l" <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a> of "`I_(2)` used `=20xx0.05=1.0` <br/> Let a and b are m" Eq of "`As_(2)O` and `As_(2)O_(5)` <a href="https://interviewquestions.tuteehub.com/tag/respectively-1186938" style="font-weight:bold;" target="_blank" title="Click to know more about RESPECTIVELY">RESPECTIVELY</a> on addition of `I_(2)` to mixture `As_(2)^(3+)` is converted `As_(2)^(5+)`. `thereforem" Eq of "As_(2)O_(3)-=m" Eq of "I_(2) used-=1.0-=m" Eq of "As_(2)^(5)` formed. <br/> `thereforea=1.0` <br/> After reaction with `I_(2)`, mixture contains all the <a href="https://interviewquestions.tuteehub.com/tag/arsenic-382124" style="font-weight:bold;" target="_blank" title="Click to know more about ARSENIC">ARSENIC</a> in `+5` oxidation state which is then titrated using `KI+` hypo. <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a>, <br/> `m" Eq of "As_(2)_(3) as As_(2)^(5+) m" Eq of "As(2)O_(5) as As^(5+)=m" Eq of "I_(2)` liberated `=m" Eq of "`hypo used <br/> `thereforea+b=(1.24)/(248)xx10^(3)` <br/> Ew of `Na_(2)S_(2)O_(5) as 5H_(2)O=(248)/(1)` <br/> `thereforea+b=5`...(ii) <br/> weight of `As_(2)O_(3)=1.0xx10^(-3)xx(198)/(4)=0.0495g` <br/> `{:(As_(2)^(3+)toAs_(2)^(5+)+4e^(-)),(Ew of As_(2)O_(3)=(198)/(4)):}` <br/> Weight of `As_(2)O_(3)=4xx10^(-3)xx(230)/(4)` <br/> `(Ew of As_(2)O_(5)=(230)/(4))` <br/> `=0.23 g` <br/> Weight of mixture`=0.0495+0.23=0.2795g`</body></html>


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