To a 25ml solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20ml of 0.3 n sodium thiosulphate solution. Calculate the volume strength of solution. (iitjeeadvanced, 1997)
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Home»Forum»Physical Chemistry»30ml of a H2O2 solution after acidification...
30ml of a H2O2 solution after acidification required 30 ml of 0.1 N KMnO4 solution for complete oxidation calculate the percentage of H2O2.
Pls explain the procedure (ans is 0.17%)
one year ago
Answers : (2)
Equivalent weight of H2O2= equivalent weight of KMnO4Equivalent weight of H2O2=0.03×0.1=0.003Wt. of H2O2= 0.003×34/2(n factor of H2O2=2) = 0.051gMass by vol. %= 0.051g/30ml ×100= 0.17%
one year ago
Gram equivalent of H2O2=Gram equivalent of KMnO4Gram equivalent of H2O2=0.03×0.1=0.003Wt. of H2O2=0.003×34/2(n factor of H2O2=2)=0.051gMass by vol.%=0.051g/30ml=0.17% By Shivam Aditya
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Home»Forum»Physical Chemistry»30ml of a H2O2 solution after acidification...
30ml of a H2O2 solution after acidification required 30 ml of 0.1 N KMnO4 solution for complete oxidation calculate the percentage of H2O2.
Pls explain the procedure (ans is 0.17%)
one year ago
Answers : (2)
Equivalent weight of H2O2= equivalent weight of KMnO4Equivalent weight of H2O2=0.03×0.1=0.003Wt. of H2O2= 0.003×34/2(n factor of H2O2=2) = 0.051gMass by vol. %= 0.051g/30ml ×100= 0.17%
one year ago
Gram equivalent of H2O2=Gram equivalent of KMnO4Gram equivalent of H2O2=0.03×0.1=0.003Wt. of H2O2=0.003×34/2(n factor of H2O2=2)=0.051gMass by vol.%=0.051g/30ml=0.17% By Shivam Aditya
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