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Q)

A solution of a salt of a metal of atomic weight $'\lambda'$ was electrolysed for 150minutes with 0.15A of current .The weight of metal deposited was 0.783g.Find X,the specific heat of metal=0.057Cal/$g^{\large\circ}C$.According to Dulong-Petit's law,approximate atomic weight of a metal $X$ specific heat=6.4

$\begin{array}{1 1}(a)\;101.86gm/mol\\(b)\;120.86gm/mol\\(c)\;96.86gm/mol\\(d)\;111.86gm/mol\end{array}$

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A)
We can use this law to calculate the approximate atomic weight of metal first.This is equal to
$\large\frac{6.4}{0.057}$$=112.28g/mol$
But this is not the value of X because $X$ is exact atomic weight.
From the data given it is possible to calculate the exact equivalent weight as
Mass of metal deposited =0.783g
Charge passed =$0.15\times 150\times 60$
$\Rightarrow 1350Coulombs$
Moles of electrons =$\large\frac{1350}{96500}$
$\Rightarrow 0.014$
$\therefore$ Equivalent weight=$\large\frac{0.783}{0.014}$
Now if we divide the approximate atomic weight,we will get the approximate valency of metal in salt.
$\therefore$ Approximate valency=$\large\frac{112.28}{55.93}$=2.007
Since valency has to be an integer,$n=2$
$\therefore $ Exact atomic weight $=55.93\times 2$
$\Rightarrow 111.86gm/mole$
Hence (d) is the correct answer.
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