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Amit Saxena Grade: upto college level
        One gram of commercial AgNO3 is dissolved in 50 mL of water. It is treated with 50 mL of KI solution. The silver iodide thus precipitated is filtered off. Excess of KI in the filtrate is titrated with (M/10) KIO3 solution in presence of 6 M HCI till all I- ions are converted into ICI. It requires solution of KI requires 30 mL of (M/10) KIO3 under similar conditions. Calculate the percentage of AgNO3 in the sample.

Reaction :
KIO3 + 2KI + 6HCI → 3ICI + 3KCI +3H2O
3 years ago

Answers : (1)

Kevin Nash
askIITians Faculty
332 Points
										Sol. The reaction is 
KIO3 + 2KI + 6HCI → 3ICI + 3KCI + 3H2O
KIO3 required for 20 mL original KI solution = 3 millimol.
⇒ 7.5 millimol KIO3 would be required for original 50 mL KI.
⇒ Original 50 mL KI solution contain 15 millimol of KI. 
After AgNO3 treatment.
5 millimol of KIO3 is required, ie, 10 millimol KI is remaining.
⇒ 5 millimol KI reacted with 5 millimol of AgNO3.
⇒ Mass of AgNO3 = 5/1000 × 170 = 0.85 g
⇒ Mass percentage of AgNO3 = 85%

3 years ago
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