What should be the weight and moles of AgCl precipitate obtained on adding 500ml of 0.20 M HCl in 30 g of $A g N O_{3}$ solution? $(A g N O_{3} = 170)$

14.35 g

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15 g

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18 g

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19 g

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Solution

The correct option is A 14.35 g
$A g N O_{3} + H C l \to A g C l + H N O_{3}$
$\frac{30}{170}$ $\frac{500 \times 0.2}{1000}$
t =0 0.176 mole 0.1 mole limiting =14.345gm
t =t 0.076 mole 0 0.1mole

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