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The CO in a 20L sample of gas was converted to `CO_2` by passing the gas over iodine pentoxide heated to `150^@C`: `I_2O_5(s)+5CO(g)to5CO_2(g)+I_2(g)` The iodine distilled at this temperature and was collected in an absorber containing 10 mL of 0.011 M `Na_2S_2O_3`.The excess hypo was back titrated with 5 mL of 0.001 M `I_2` solution. What must be the milligrams of Co in 1 L of the original gas sample ?A. 0.35 mgB. 0.70 mgC. 0.07 mgD. 3.5 mg |
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Answer» Correct Answer - A meq of `l_2` liberated =10 x 0.011 x 1-5 x 0.001 x 2=0.1 Millimoles of CO=`0.1/2 xx5 =0.25` Weight of CO in 1L =`0.25/20xx28 =0.35 mg` |
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