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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)

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5 years ago

Ashutosh Sharma
181 Points
```							what’s the requirement of question?
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5 years ago
Mahesh Sharma
32 Points
```							According to the question      wt of metal obtained on heating 4.77gmof oxide B=3.81gmwt of oxygen lost=wt of oxide --wt of metal                              =4.77-3.81                             =0.96gmEquivalent  mass of metal in oxide B=(wt of metal /wt of metal )*8                                                          =3.81*8/0.96                                                          =31.50now valency of metal in B=atomic mass of metal/equivalent wt                                         =63.50/31.5=2thus formula of black metal oxide = M+2+O-2=MONow 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=0.48gmEquivalent wt of metal in A =(3.81/0.48)*8=63.50Valency of  metal A= Atomic wt /  Equivalent wt                                 =63.50/63.50=1Formula of oxide A =M+ + O-2=M2​O
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4 years ago
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