How much amount of CuSO4- 5H2O is required for liberation of 2-54 g of I2 when titrated with KI-

The correct option is C 4.99 g 2CuSO_4· 5H_2O + 4KI→ Cu_2I_2 + 2K_2SO_4 + 254 gI_2 + 10H_2O Molecular weight of 2CuSO_4· 5H_2O =[2(63.5 + 3 ...

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Standard XII

Chemistry

Stoichiometric Calculations

How much amou...

Question

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

2.5 g

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4.99 g

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2.4 g

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1.2 g

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Solution

The correct option is C $4.99 g$
$2 C u S O_{4} \cdot 5 H_{2} O + 4 K I \to C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2} 254 g + 10 H_{2} O$

Molecular weight of $2 C u S O_{4} \cdot 5 H_{2} O$ $= [2 (63.5 + 32 + 64) + 10 (18)] g = 499 g$

$254 g$ of $I_{2}$ is liberated by $499 g$ $C u S O_{4} \cdot 5 H_{2} O$ .

$2.54 g$ of $I_{2}$ will be liberated by $x g C u S O_{4} \cdot 5 H_{2} O$ .

$x = \frac{499 \times 2.54}{254} = 4.99 g$

Hence, the correct option is $B$

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Similar questions

Q. How much amount of

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

required for liberation of

2.54

I_{2}

when titrated with

K I

Q. 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

638.0

g of

C u S O_{4}

solution is titrated with excess of

0.2 M

K I

solution. The liberated

I_{2}

required

400

mL of

1.0 M

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

for complete reaction. The percentage purity of

C u S O_{4}

in the sample is:

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

to prepare

100

ml of

0.02 M C u S O_{4} (a q)

10

ml of

0.02 M

solution of

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

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

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

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

) which requires

V

ml of

0.1 M

hypo solution for its complete reduction.The amount of

I_{2}

liberated in the reaction of

10

ml of

0.02 M

solution with excess

K I

is :

Q. 3 gm of

{C u S O}_{4}

5 H_{2} O

sample containing inert impurity reacted with KI and the liberated

I_{2}

reacted with

10 m l

0.5 M

{N a}_{2} S_{2} O_{3}

. The % purity of

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

will be:

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How much amount of CuSO4⋅5H2O is required for liberation of 2.54 g of I2 when titrated with KI?

The correct option is C 4.99 g2CuSO4⋅5H2O+4KI→Cu2I2+2K2SO4+I2254 g+10H2OMolecular weight of 2CuSO4⋅5H2O=[2(63.5+32+64)+10(18)]g=499 g254 g of I2 is liberated by 499g CuSO4⋅5H2O.2.54 g of I2 will be liberated by x g CuSO4⋅5H2O.x=499×2.54254=4.99 gHence, the correct option is B

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