11Solution?
90 gm of a metal reacts with 80 gm of oxygen to form metal oxide. The metal oxide requires 20 litre of 1/3 N of oxidizing agent to oxidise to M5+. The molecular weight of the metal is
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1manish is it really neccessary to assume the molecular formulae of the metal oxide to be M2Ox type only i am getting the same ans by taking MxOy type i think question dont state that ....
1357M +O2→M2Ox→M5+
From definition, equivalent weight is defined as the weight of metal which combines with 8 g of oxygen.
so the equivalent weight of metal is 90/10=9
grameq. of metal=10
gram eq. of metaloxide in first reaction is 10
gram eq. of metaloxide in second reaction is 20/3
But moles of metaloxide will remain same in first as well as second reaction
n1=n2
geq/x1=geq/x2
10/(2*x)=(20/3)/(2*(5-x))
x=3
M=E*x=9*3=27
1yaar tumlog subjective me kaise tukka marte ho ?
how i proceeded..
1) found no of equi (20/3)
2) M(90) + O(80) ---> MO(170)
so 80g has to 20/3 eq.
so, no of moles = 80/32 = 5/2 = 20/3 X n-factor
=> n = 3/8
now, 90/ M = 3/8 X 20/3 => M=36 g
is this the ans?
33half waht aragorn guesssed and most near to me :P
but this was my first answer which i gave of chemsitry question after solving (in a test or here )
and got it wrong... :(
33So its hypothetical but how u replied before my question while i was writing the post u replied.. :)
33Me getting 21.5 :(
but this should not be as there is no such metal :(
11Yup. Saw +5, took it as Mn5+ and saw molecular mass of Mn and gave it as 55 [4]