You visited us 296 times! Enjoying our articles? Unlock Full Access!

Electronic Configuration

What is the i...

Question

What is the increasing order of equivalent weights of the following oxidising agents?
(i) $K M n O_{4} ⟶ M n^{2 +}$
(ii) $K_{2} C r_{2} O_{7} ⟶ C r^{3 +}$
(iii) $K C l O_{3} ⟶ C l^{-}$
[Atomic wt.:
$K = 39, C r = 52, M n = 55, C l = 35.5, O = 16$ ]

K C l O_{3} < K M n O_{4} < K_{2} C r_{2} O_{7}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

K_{2} C r_{2} O_{7} < K M n O_{4} < K C l O_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

K M n O_{4} < K_{2} C r_{2} O_{7} < K C l O_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

K C l O_{3} < K_{2} C r_{2} O_{7} < K M n O_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $K C l O_{3} < K M n O_{4} < K_{2} C r_{2} O_{7}$
$(i) K 7 + M n O_{4} ⟶ M n^{2 +}$ $\Rightarrow E_{1} = \frac{39 + 55 + 64}{5} = 31.6$
$(i i) K_{2} 6 + C r_{2} O_{7} ⟶ C r^{3 +}$ $\Rightarrow E_{2} = \frac{39 \times 2 + 52 \times 2 + 16 \times 7}{6} = 49$
$(i i i) K 5 + C l O_{3} ⟶ C l^{-}$
$\Rightarrow E_{3} = \frac{39 + 35.5 + 48}{6} = 20.4$

Suggest Corrections

Similar questions

K M n O_{4} ⟶ M n^{2 +}

K_{2} C r_{2} O_{7} ⟶ C r^{3 +}

K C l O_{3} ⟶ C l^{-}

K = 39, C r = 52, M n = 55, C l = 35.5, O = 16

K M n O_{4} ⟶ M n^{2 +}

K_{2} C r_{2} O_{7} ⟶ C r^{3 +}

K C l O_{3} ⟶ C l^{-}

K = 39, C r = 52, M n = 55, C l = 35.5, O = 16

a) What is percentage composition of potassium chlorate (KClO₃).

Given:Atomic mass Of K=39,Cl=35.5andO=16

K M n O_{4}

40 g

K = 39; M n = 55; O = 16

20

K C l O_{3}

[K = 39, C l = 35.5, O = 16]

Join BYJU'S Learning Program