The ion A n- is oxidised to AO 3 - by Mn O 4 - changing to Mn 2- in acidic solution- Given that 2-68- 10 -3 mol pf A n- requires 1-61- 10 -3 mol of Mn O 4 - The value of n is -

The correct option is A 2 +n A ^ n+ ⟶ +5 AO _ 3 ^ (5 n) change +7 Mn O _ 4 ^ ⟶ +2 Mn ^ 2+ 5 change Balancing oxidation number, ...

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Byju's Answer

Standard XII

Chemistry

Concentration Terms

The ion A n...

Question

The ion $A^{n +}$ is oxidised to ${A O}_{3}^{-}$ by $M n O_{4}^{-}$ 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 $n$ is :

2

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Solution

The correct option is A $2$
$A^{n +} + n ⟶ {A O}_{3}^{-} + 5 c h a n g e (5 - n)$
$M n O_{4}^{-} + 7 ⟶ {M n}^{2 +} + 2 c h a n g e 5$
Balancing oxidation number,
$(5 - n) M n O_{4}^{-} + 5 A^{n +} ⟶ 5 {A O}_{3}^{-} + (5 - n) {M n}^{2 +}$
Thus, $\frac{m o l e s o f M n O_{4}}{m o l e s o f A^{n +}} = \frac{1.61 \times 10^{- 3}}{2.68 \times 10^{- 3}}$
$∴ \frac{5 - n}{5} = \frac{1.61 \times 10^{- 3}}{2.68 \times 10^{- 3}}$
$(5 - n) = 3$
$∴ n = 2$

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Similar questions

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}

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

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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

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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

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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

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?

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The ion An+ is oxidised to AO−3 by MnO−4 changing to Mn2+ in acidic solution. Given that 2.68×10−3 mol pf An+ requires 1.61×10−3 mol of MnO−4. The value of n is :

The correct option is A 2An++n⟶AO−3+5change(5−n)MnO−4+7⟶Mn2++2change5Balancing oxidation number,(5−n)MnO−4+5An+⟶5AO−3+(5−n)Mn2+Thus, moles of MnO4moles of An+=1.61×10−32.68×10−3∴5−n5=1.61×10−32.68×10−3(5−n)=3∴n=2

The ion $A^{n +}$ is oxidised to ${A O}_{3}^{-}$ by $M n O_{4}^{-}$ 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 $n$ is :