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1.878 g of MBr(X) when heated in a strea...

`1.878 g` of `MBr_(X)` when heated in a stream of `HCl` gas was completely converted to chloride `MCl_(X)` which weighed `1.0 g` The specific heat of metal is `0.14 cal g^(-1)`. Calculate the molecular weight of the metal bromide.

Text Solution

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The equation for the reaction given in the question will be `MBr_(x)+HClrarrMCl_(x)+xHBr`
From Dulong's and Petit's law we know that the atomic mass of the metal `=(6.4)/(0.14)="45.71g/mol"`
Now, `45.71+ 80x` metal bromide gives `45.71+ 35.5x` metal chloride . So, `(1.878)/(45.71+80x)=(1)/(45.71+35.5x)`.
Solving this equation we get : `x = 3`. Molecular mass of metal bromide `45.71+80(3)=285.71"g/mole"`
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