Question

The weight of nascent oxygen in milligrams obtained from 6.32 g of potassium
permanganate (Molecular weight 158) in acid medium is

Answer 1

Answer:The Mn in KMnO4 exists in +7 state. In Acidic medium, This Mn +7 goes to Mn+2 state and hence there is a net gain of 5 electrons. Therefore,The Equivalent Weight = 158/5 = 31.6 grams.

Explanation:

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Answer 2

Answer:

The weight of nascent oxygen obtained is 1600mg

Explanation:

Given that,

Mass of potassium permanganate used in the reaction is 6.32gm.

We are required to find the number of moles of nascent oxygen obtained and for that we shall write the chemical equation of potassium permanganate in acidic medium in which nascent oxygen is obtained as a product of the reaction.The chemical equation is

$\\ 2KMnO_{4}+3H_{2}SO_{4}-->K_{2}SO_{4}+2MnSO_{4}+3H_{2}O+5[O]$

From the above chemical reaction,we can interpret that 2 moles of Potassium permanganate gives 5 moles of nascent oxygen.

We are required to find the no.of moles of potassium permanganate using its molecular weight(158g/mol)

$n = \frac{6.32}{158} = 0.04 \: mol$

The no.of moles of nascent oxygen which 0.04 moles of Permanganate gives is

$\frac{0.04}{2} \times 5 = 0.1 \: mol$

Therefore,weight of 0.1 mole of nascent oxygen is

$0.1 \times 16 = 1.6 \: gm = 1600mg$

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