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Standard XII

Chemistry

Stoichiometric Calculations

x g of Ag w...

Question

$x$ g of $A g$ was dissolved in $H N O_{3}$ and the solution was treated with excess of $N a C l$ . When $2.87$ g of $A g C l$ was precipitated, the value of $x$ is :

1.08

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2.16

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2.70

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1.62

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Solution

The correct option is D $2.16$
Equivalents of Ag $=$ Equivalents of AgCl

$\frac{x}{108} = \frac{2.87}{[108 + 35.5]}$

$x = 2.16$ g

Hence, the value of $x$ is $2.16$ .

Hence, the correct option is $B$

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x g of Ag was dissolved in HNO3 and the solution was treated with excess of NaCl. When 2.87 g of AgCl was precipitated, the value of x is :

The correct option is D 2.16Equivalents of Ag = Equivalents of AgClx108=2.87[108+35.5]x=2.16 gHence, the value of x is 2.16.Hence, the correct option is B

$x$ g of $A g$ was dissolved in $H N O_{3}$ and the solution was treated with excess of $N a C l$ . When $2.87$ g of $A g C l$ was precipitated, the value of $x$ is :

The correct option is D $2.16$
Equivalents of Ag $=$ Equivalents of AgCl

$\frac{x}{108} = \frac{2.87}{[108 + 35.5]}$

$x = 2.16$ g

Hence, the value of $x$ is $2.16$ .

Hence, the correct option is $B$