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Q) 3.55 gm of bleaching powder is suspended in water and treated with dilute H2SO4 and excess of potassium iodide liberated iodine just requires 80 ml of iodine solution calculate percentage av of chlorine in bleaching powder.

Asked by Anish | 01 May, 2019, 13:27: PM

Expert Answer

Question:

3.55 g sample of bleaching powder suspended in H₂ O was treated with dil.H₂SO₄ and KI solution. Iodine thus liberated requires 80 mL of 0.2 M hypo for titration. Calculate the % of available chlorine in bleaching powder.

Answer:

Bleaching powder + Acid + KI → KI3 + starch (hypo) → end point (Blue →colourless)

Formula:

percent sign space of space Cl subscript 2 space space equals space end subscript fraction numerator 3.55 space cross times Molarity space of space hypo space solution space cross times space mL space of space hypo space solution space used space in space titration over denominator straight W space left parenthesis in space gm right parenthesis end fraction Moles space of space iodine space equals space moles space of space chlorine space equals space fraction numerator 80 cross times 0.2 over denominator 2 end fraction cross times 10 to the power of negative 3 end exponent equals space 8 space cross times 10 to the power of negative 3 end exponent hence space required space percentage space equals space fraction numerator 8 space cross times space 71 space cross times space 10 to the power of negative 3 end exponent over denominator 3.55 end fraction space cross times space 100 space equals space 16 space percent sign

Answered by Ramandeep | 02 May, 2019, 13:01: PM

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