$x$ g of $Ag$ was dissolved in $HNO_{3}$ and the solution was treated with excess of $NaCl$ when 2.87 g of $AgCl$ was precipitated. The value of $x$ is

Question

Dr. Abubacker · Accepted Answer

(b)
$2Ag + 2HNO_{3} \longrightarrow 2AgNO_{3} + H_{2}$
$2AgNO_{3} + 2NaCl \longrightarrow 2AgCl + NaNO_{3}$
$AgCl \equiv AgNO_{3} \equiv Ag$
143.5g 170 g 108g
$\because 143.5\ g\ \ \ \ AgCl$ is obtained from $Ag = 108g$
$	herefore 2.87\ g\ AgCl$ is obtained from $Ag = \frac{108 	imes 2.87}{143.5}$
$= 2.16g$

Q. $x$ g of $Ag$ was dissolved in $HNO_{3}$ and the solution was treated with excess of $NaCl$ when 2.87 g of $AgCl$ was precipitated. The value of $x$ is

Explanation

Reviewed by Dr. Abubacker

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