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Byju's Answer

Standard XII

Chemistry

Central Atom and Ligands

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 g

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2.17 g

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2.70 g

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1.62 g

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Solution

The correct option is D $2.17 g$
$A g + C l ⟶ A g C l$
Atomic mass of $A g = 108$ g
Atomic mass of $C l = 35.5$ g
Molecular mass of $A g C l = 143.5$ g
143.5 g $A g C l$ is obtained from 108 g $A g$
2.87 g $A g C l$ is obtained from $= \frac{108 \times 2.87}{143.5} = 2.17 g$

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Similar questions

Q. A

100 m L

solution containing

A g N O_{3}

was treated with excess

N a C l

to completely precipitate the silver as

A g C l

. If

5.7 g A g C l

was obtained, what was the concentration of

A g^{+}

in the original solution? (At. mass of

A g = 108 g / m o l

)

Q. A 100 -milliliter solution of

A g N O_{3}

and excess

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were treated and the silver was precipitate as

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. If 5.7 g

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was obtained, what was the concentration of Ag

^{+}

in the original solution?

Q. A

100

ml solution containing

A g N O_{3}

was treated with excess

N a C l

to completely precipitate the silver as

A g C l

. If

5.7

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was obtained, what was the concentration of

A g^{+}

in the original solution?

Q. The aqueous solution of inorganic compound

(X)

yielded a white precipitate when treated with dilute

H N O_{3}

and

A g N O_{3}

. Another sample of the solution of

(X)

when treated with

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gave a white precipitate first which dissolved in excess of

N a O H

yielding a colorless solution. When

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Identify the compound

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Q. About

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is dissolved in water and then treated with excess of silver nitrate solution. The weight of silver chloride precipitated was found to be

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

The correct option is D 2.17gAg+Cl⟶AgClAtomic mass of Ag=108 gAtomic mass of Cl=35.5 gMolecular mass of AgCl=143.5 g143.5 g AgCl is obtained from 108 g Ag2.87 g AgCl is obtained from =108×2.87143.5=2.17 g

$x$ 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 $2.17 g$
$A g + C l ⟶ A g C l$
Atomic mass of $A g = 108$ g
Atomic mass of $C l = 35.5$ g
Molecular mass of $A g C l = 143.5$ g
143.5 g $A g C l$ is obtained from 108 g $A g$
2.87 g $A g C l$ is obtained from $= \frac{108 \times 2.87}{143.5} = 2.17 g$