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

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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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