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Density

How much amou...

Question

How much amount of $C u S O_{4} .5 H_{2} O$ required for liberation of $2.54 g I_{2}$ when titrated with $K I$ ?

2.5 g m

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4.99 g m

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2.4 g m

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1.2 g m

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Solution

The correct option is B $4.99 g m$
$2 C u S O_{4} + 4 K I \to {C u}_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}$

Molecular wt of $C u S O_{4} .5 H_{2} O$ = 249.6 g/mol

Molecular wt of $I_{2}$ = 253.8 g/mol

2 moles (500 g) of $C u S O_{4} .5 H_{2} O$ is required to liberate 1 mol (254 g) of $I_{2}$

Amount of $C u S O_{4} .5 H_{2} O$ required for $2.54 g$ of $I_{2}$ is

$= 2.54 \times \frac{500}{254} = 5$

$5 g C u S O_{4} .5 H_{2} O$

Therefore, option B is correct.

Suggest Corrections

Similar questions

C u S O_{4} . 5 H_{2} O

2.54

I_{2}

K I

C u S O_{4} \cdot 5 H_{2} O

2.54 g

I_{2}

K I

0.1 g K I O_{3}

K I

H C l

45 m L

(a)

C u S O_{4}

K I

I_{2}

2 C u S O_{4} + 4 K I ⟶ C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}

(b)

H g_{5} (I O_{6})_{2}

K I

H C l

H g_{5} (I O_{6})_{2} + 34 K I + 24 H C l ⟶ 5 K_{2} H g I_{4} + 8 I_{2} + 24 K C l + 12 H_{2} O

(c)

N a_{2} S_{2} O_{3}

0.0499

C u S O_{4} .5 H_{2} O

N a_{2} S_{2} O_{3}

I_{2}

0.7245

H g_{5} (I O_{6})_{2}

C u S O_{4} .5 H_{2} O = 249.5

H g_{5} (I O_{6})_{2} = 1448.5

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