23u can use dis also
no of equivalents of ABrx= no of eq of AgBr = 1.32/188
i.e
0.722E+80 = 1.32188
A = element,x = its valency , E = its eq wt
so E + 80 = 102.83
E = 91.2/x = 102.83-80=22.83
so x = 91.2/22.83≈4
0.722 g of bromide of an element A having atomic weight 91.2 in a series of reaction gives 1.32 g of AgBr. Find the valency of element A.
Also,show the steps...
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2 Answers
1let the valency of the element "E" be 'x'
One of the reaction in the series will be ::
E-Brx + Ag -> Ag-Br + E-Brx-1 .... (1)
mass of AgBr is 1.32 g
so no. of moles of AgBr is 1.32188
where 188 is the molecular mass of AgBr
this gives 0.007 as the no. of moles so eqn.. (1) bcomes ::
0.007 E-Brx + 0.007 Ag ->0.007 Ag-Br+other product...... (2)
total wt. of the bromide i.e E-Brx acc. to the eqn (2) will be ::
0.007[atomic mass of E + x (atomic mass of Br)]
this value is equal to the given mass of bromide i.e. : 0.772gm
=> 0.007[91.2+ 'x'80]= 0.772 which gives x = 1/4
... so the valency of the element 'E' is 4
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