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Grade: 11
        
On being heated in oxygen 5.72 g of red metallic oxide A was converted to 6.36 g black metallic oxide B. When 4.77 g of B was heated in a stream of H2 gas, 3.81 g of metal M was formed.(Given, atomic weight of metal is 63.60)
3 years ago

Answers : (2)

Ashutosh Sharma
askIITians Faculty
181 Points
							what’s the requirement of question?
3 years ago
Mahesh Sharma
32 Points
							
According to the question 
   
  wt of metal obtained on heating 4.77gmof oxide B=3.81gm

wt of oxygen lost=wt of oxide --wt of metal 
                             =4.77-3.81
                             =0.96gm
Equivalent  mass of metal in oxide B=(wt of metal /wt of metal )*8

                                                          =3.81*8/0.96
                                                          =31.50

now valency of metal in B=atomic mass of metal/equivalent wt
                                         =63.50/31.5=2
thus formula of black metal oxide = M+2+O-2=MO

Now since according to the first Question .......

6.36gm Of black oxide obtained by 5.72gm of A(Red Metallic oxide )

4.77gm of B is obtained by (5.72/6.36)*4.77gm A =4.29gm A 

Now we can say that,

3.81gm M obtained by 4.29gm A 

wt of oxygen lost form A = wt of A --wt of metal 

                                     =4.29-3.81=0.48gm

Equivalent wt of metal in A =(3.81/0.48)*8=63.50
Valency of  metal A= Atomic wt /  Equivalent wt 
                                =63.50/63.50=1
Formula of oxide A =M+ + O-2=M2​O
one year ago
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