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Q.

For thereaction, $M^{+x}+MnO_4^-rightarrow MO_3^-+Mn^{+2}+frac{1}{2}O_2$ , if one mole of $MnO_4^-$ oxidises 1.67 moles of M^x+ to MO₃^- , then the value of 'x' in the reaction is

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a

5

b

3

c

2

d

1

answer is C.

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Detailed Solution

$large {M^{{ +x}}}; + ;MnO_4^ - ; to ;MO_3^ - ; + ;M{n^{ + 2}}; + ;frac{1}{2}{O_2}$
$begin{array}{*{20}{c}} {RHR} \ {mathop {MnO_4^ - }limits_{ + 7} ; to ;mathop {M{n^{ + 2}}}limits_{ + 2} } \ {Decrease;in;Ox.state/mole; = ;5} end{array}left| {begin{array}{*{20}{c}} {OHR} \ {mathop {{M^{ + x}}}limits_{ + x} ; to ;mathop {MO_3^ - }limits_{ + 5} } \ {Increase;in;Ox.state/;mole;of;{M^{ + x}}; = ;2} end{array}} right.$
Increase in Ox.state for 1.67 mole of M^+x = (5-x) 1.67
Total decrease in Ox.state = total increase in Ox.state
5 = (5-x)1.67
5 = 8.35 - 1.67x
1.67x = 3.35
X = 2

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