1st chapter of chem...

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...

2 Answers

let the valency of the element "E" be 'x'

One of the reaction in the series will be ::

E-Br_x + Ag -> Ag-Br + E-Br_x-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-Br_x + 0.007 Ag ->0.007 Ag-Br+other product...... (2)

total wt. of the bromide i.e E-Br_x 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

u can use dis also

no of equivalents of ABr_x= 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 = 91.2/22.83â‰ˆ4

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