KMnO4 oxidizes XOn- ion to X3- ion- itself changing to Mn2- in acid medium- 2-68- 10-3 mole of Xn-requires 1-61- 10-3 mole of MnO4- If the weight of 1g equivalent of XCIn is 56- then the atomic mass of X is -

The correct option is A 41 KMnO_4 + nbsp; X^+n→ XO_3^ +Mn^+2 1.61× 10^ 3 moles of MnO_4 reacts with nbsp; 2.63× 10^ 3 moles ...

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Byju's Answer

Standard XII

Chemistry

Lewis Theory

KMnO4 oxidize...

Question

$K M n O_{4}$ 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 $X C I_{n}$ is $56$ , then the atomic mass of X is :

41

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Solution

The correct option is A $41$
$K M n O_{4}$ + $X^{+ n} \to X O_{3}^{-} + M n^{+ 2}$
$1.61 \times 10^{- 3}$ moles of $M n O_{4}$ reacts with $2.63 \times 10^{- 3}$ moles of $X^{+ n}$
No. of Equivalents of $K M n O_{4}$ = No. of Equivalents of $X^{+ n}$
$1.61 \times 10^{- 3} \times 5 = 2.63 \times 10^{- 3} \times (5 - n)$
$n = 2 \Rightarrow 56 = \frac{M}{2} + 35.5$ $M = 41$ g

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Similar questions

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}

56

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

2.68 \times 10^{- 3}

moles of a solution containing an ion

A^{n +}

require

1.61 \times 10^{- 3}

moles

M n O_{4}^{-}

for the oxidation of

A^{n +}

A O_{3}^{-}

in acid medium. What is the value of n?

2.68 \times 10^{- 3}

moles of a solution containing an ion

A^{n +}

require

1.61 \times 10^{- 3}

moles of

M n O_{4}^{-}

for the oxidation of

A^{n +}

A O_{3}^{-}

in acidic medium. What is the value of

n

Q. The ion

A^{n +}

is oxidised to

A O_{3}^{-}

M n O_{4}^{-}

changing to

M n^{2 +}

in acid medium. Given that

2.68 \times 10^{- 3}

mole of

A^{n +}

required

1.61 \times 10^{- 3}

mole

M n O^{-}

. What is the value of n?

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KMnO4 oxidizes XOn+ ion to X3− ion, itself changing to Mn2+ in acid medium. 2.68×10−3 mole of Xn+requires 1.61×10−3 mole of MnO4−. If the weight of 1g equivalent of XCIn is 56, then the atomic mass of X is :

The correct option is A 41KMnO4 + X+n→XO−3+Mn+21.61×10−3 moles of MnO4 reacts with 2.63×10−3 moles of X+nNo. of Equivalents of KMnO4= No. of Equivalents ofX+n1.61×10−3×5=2.63×10−3×(5−n)n=2⇒56=M2+35.5 M=41 g