thanx thanx thanx dude ....
1g sample commercial AgNO3 is dissolved in 50ml of water. It is treated with 50ml of KI solution. The silver iodide thus precipated is filtered off. Excess of KI in the filtrate is titrated with M/10 KIO3 solution in the presence of 6M hCl till all the I- ions are converted into ICl. It requires 50ml of M/10 KIO3 solution. 20ml of same stock solution of KI requires 30ml of M/10 KIO3 solution under similar condition Calculate % of Silver Nitrate in the sample
.KIO3+2KI+6HCL→3ICl+3KCl+3H20
PLEASE FRNDS URGENTLY SOLVE THIS RATES ASSURED ??????
PLEASE DUDE SOLVE THIS AND PLZ GIVE STEPS ALSO THE ANSWER IS 85 %
I DESPERATELY HOPE AN ANSWER FROM YOU ............
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http://targetiit.com/iit_jee_forum/posts/stoichiometry_3440.html