KMn O 4 oxidises Xn-1 ion into X O 3 - itself changing to Mn 2- in acid solution- 2-68- 10 -3 mol of Xn-1 requires 1-61- 10 -3 mol of Mn O 4 - What is atomic weight of X- if the weight of 1 g equivalent of XCln is 56-

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Byju's Answer

Standard XII

Chemistry

Percentage Composition

KMn O 4 oxidi...

Question

$K M n O_{4}$ oxidises $X^{n + 1}$ ion into ${X O}_{3}^{-}$ , itself changing to ${M n}^{2 +}$ in acid solution. $2.68 \times 10^{- 3}$ mol of $X^{n + 1}$ requires $1.61 \times 10^{- 3}$ mol of $M n O_{4}^{-}$ . What is atomic weight of $X$ , if the weight of $1$ g equivalent of ${X C l}_{n}$ is $56$ ?

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Solution

Here, $n = 1$
The oxidation number changes from +2 to +5. The change in the oxidation number is 3.
$X^{2 +} + 2 ⟶ X O_{3}^{-} + 5 c h a n g e 3$
The expression for the equivalent mass is as given below.
$E q u i v a l e n t m a s s o f {X C l}_{2} = \frac{M}{3} = 56$
$∴ M = 56 \times 3 = 168$
${X C l}_{2} = M + 35.5 \times 2 = 168$
$∴ M = 97$
Hence, the atomic weight of X is 97.

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Similar questions

K M n O_{4}

oxidizes

X O^{n +}

ion to

X {_{3}}^{-}

ion, itself changing to

M n^{2 +}

in acid medium.

2.68 \times 10^{- 3}

mole of

X^{n +}

requires

1.61 \times 10^{- 3}

mole of

M n O {_{4}}^{-}

. If the weight of

1

g equivalent of

X C I_{n}

56

, then the atomic mass of X is :

Q. The ion

A^{n +}

is oxidised to

{A O_{3}}^{-}

{M n O_{4}}^{-}

changing to

M n^{2 +}

in acid solution. Given that

2.68 \times 10^{- 3}

mol of

A^{n +}

requires

1.61 \times 10^{- 3}

mol of

{M n O_{4}}^{-}

.
What is the weight of one gram equivalent of

A C l_{n}

for the above reaction if the atomic mass of

A

97

Q. The ion

A^{n +}

is oxidised to

{A O}_{3}^{-}

M n O_{4}^{-}

changing to

{M n}^{2 +}

in acidic solution. Given that

2.68 \times 10^{- 3}

mol pf

A^{n +}

requires

1.61 \times 10^{- 3}

mol of

M n O_{4}^{-}

. The value of

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is :

Q. A solution containing

2.68 \times 10^{- 3} mol

A^{n +}

ions requires

1.61 \times 10^{- 3} mol

M n O_{4}^{-}

for the complete oxidation of

A^{n +}

A O_{3}^{-}

in acidic medium. What is the value of

n

Q. Assertion :The equivalent weight of

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31.6

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KMnO4 oxidises Xn+1 ion into XO−3, itself changing to Mn2+ in acid solution. 2.68×10−3 mol of Xn+1 requires 1.61×10−3 mol of MnO−4. What is atomic weight of X, if the weight of 1 g equivalent of XCln is 56?

Here, n=1The oxidation number changes from +2 to +5. The change in the oxidation number is 3.X2++2⟶XO−3+5change3The expression for the equivalent mass is as given below.Equivalent mass of XCl2=M3=56∴M=56×3=168XCl2=M+35.5×2=168∴M=97Hence, the atomic weight of X is 97.