Question

The CO in a 28.9-L sample of gas was converted to CO2 by passing the sample over 1205 heated to 150°C 120g(s) + 5 CO(g) +5CO2() + 12(9) The lodine was distilled at this temperature and was collected in an absorber containing 9.03 mL of 0.01567 M Na2S2O3. 12(g) +25203 - (aq) + 21"(aq) +S40% (aq) The excess Na2S203 was back-titrated with 2.5 mL of 0.0082 M 12 solution. Calculate the concentration of CO (28.01 g/mol) in mg per liter of sample.

Answer 1

Iz (g) + 25.02- laqu 21 lag + 540% - (ap) Given Concentocation of I solution = c = 0.0082 M. Volume of I solution = v= 2.5mL. No of m.moles of Iz = CXV = 0.0082m x 2.5ml = 0.0205 mmole 105 (5) + 5 coug) - 560,'g) + Ig) | 5 mole of colg reacts with Imole of Ila) according Imole of I2 Og(s) to give to balanced reaction.

No of moles of colg reacted = 5 x 0.020sm-mol 0.1025m.mole Molar mass of cold) = 28.018 mole Amount of colg Present = 0.1025 x 10 molex 28.019 male 2.87X108 19 = 10²mg, Imol = 10 m.mole Amount of colg) Present = 2.87 mg r = 2.5L = 0.0 Concentration of colg) - 2.8 mg - tisch 1148 mg/ 0.0025L Concentration of colg) = 1148mg/L Answes 2.8 Concentration SS 1148 m