1 g sample of AgNO3 is dissolved in 50 mL of water- It is titrated with 50 mL of KI solution- The AgI precipitated is filtered off- Excess of Kl in filtrate is titrated with M-10 KIO3 in presence of 6 M HCl till all I- converted into ICl- It requires 50 mL of M-10 KIO3 solution- 20 mL of the same stock solution of KI requires 30 mL of M-10 KIO3 under similar conditions- Calculate the percentage of AgNO3 in sample- The reaction is given below- KIO3-2KI-6HCl-3ICl-3KCl-3H2O

The correct option is A 85% 20 mL of the same stock solution of KI requires 30 mL of M/10 KIO_3 .50 mL of the same stock solution of KI requires ...

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Standard XII

Chemistry

Stoichiometric Calculations

1 g sample of...

Question

$1$ g sample of $A g N O_{3}$ is dissolved in $50$ mL of water. It is titrated with $50$ mL of $K I$ solution. The $A g I$ precipitated is filtered off. Excess of $K l$ in filtrate is titrated with $\frac{M}{10}$ $K I O_{3}$ in presence of $6 M H C l$ till all $I^{-}$ converted into $I C l$ . It requires $50$ mL of $\frac{M}{10}$ $K I O_{3}$ solution. $20$ mL of the same stock solution of $K I$ requires $30$ mL of $\frac{M}{10}$ $K I O_{3}$ under similar conditions. Calculate the percentage of $A g N O_{3}$ in sample. The reaction is given below:
$K I O_{3} + 2 K I + 6 H C l ⟶ 3 I C l + 3 K C l + 3 H_{2} O$

85 %

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75 %

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50 %

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Solution

The correct option is A $85 %$
20 mL of the same stock solution of KI requires 30 mL of $\frac{M}{10} K I O_{3}$ .
50 mL of the same stock solution of KI requires $30 \times \frac{50}{20} = 75$ mL of $\frac{M}{10} K I O_{3}$ .
Out of this, 50 mL is required for reaction with excess KI.
Hence, $75 - 50 = 25$ mL of $K I O_{3}$ corresponds to the KI that has reacted with $A g N O_{3}$ .
1 mole $K I O_{3} = 2$ mole KI $= 2 m o l e A g N O_{3}$
Number of moles of $A g N O_{3} = 2 \times \frac{25}{10 \times 1000} = 0.005$ moles
Mass of $A g N O_{3} = 0.005 \times 169.87 = 0.85$ g
The percentage of silver nitrate in the sample $= \frac{0.85}{1} \times 100 = 85$ %.

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