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Percentage Composition

A sample cont...

Question

A sample containing $C a {B r}_{2}$ and $N a I$ in mass ratio $\frac{2}{9}$ was dissolved in water and treated with sufficient amount of aq. $A g {N O}_{3}$ solution to form a mixed precipitate of $A g I$ and $A g B r$ . The weight of mixed precipitate was found to be $1786 g m$ .
Select the correct option.

Moles of

C a {B r}_{2}

and

N a I

in the original sample was

1

and

6

respectively

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Mass of original sample was

900 g m

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Mass of

C a {B r}_{2}

in the original sample was

400 g m

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Numer of moles of

A g B r

produced is

1

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Solution

The correct option is A Moles of $C a {B r}_{2}$ and $N a I$ in the original sample was $1$ and $6$ respectively
Given:- ${C a B r}_{2}$ and NaI in mass ratio $\frac{2}{9}$ treated with ${A g N O}_{3}$ .
So, ${C a B r}_{2} + 2 A g + O_{3} ⟶ 2 A g B r + C a {({N O}_{3})}_{2}$
also $N a I + A g N O \to A g I + {N a N O}_{3}$
Given weight of mixed precipitate $= 1786$ gram
$(A g I + A g B r)$
If we take initial mass of mixture $= u$
then $C a B r_{2} = \frac{2 x}{11}$ and $N a I = \frac{9 u}{11}$
moles of ${C a B r}_{2} = \frac{2 x}{11 \times 200},$ moles of $N a I = \frac{9 x}{11 \times 150}$
So, After. reaction moles of $A g B r = \frac{4 x}{11 \times 200}$
moles of $A g I = \frac{9 x}{11 \times 150}$
mass of mixed precipitate $= \frac{4 x}{11 \times 200} \times 188 + \frac{9 x}{11 \times 150} \times 235$
givan, $\frac{4 x \times 188}{11 \times 200} + \frac{9 x}{11 \times 150} \times 235 = 178.6$ gram
So, $x = 1100$ gram $, N o w$
$M a s s o f C a B r_{2} = \frac{2 \times 1100}{11 \times 200} = 1$
$M a s s o f N a I = \frac{9 \times 1100}{11 \times 150} = 6$
So, $A$ is correct answer

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A sample containing CaBr2 and NaI in mass ratio 29 was dissolved in water and treated with sufficient amount of aq. AgNO3 solution to form a mixed precipitate of AgI and AgBr. The weight of mixed precipitate was found to be 1786gm.Select the correct option.