A sample of a metal chloride weighing $0.22 g$ required $0.51 g$ of $A g N O_{3}$ to precipitate the chloride completely. The specific heat of the metal is $0.057$ . Find out the molecular formula of the chloride if the symbol of the metal is $^{'} M^{'}$ .
$[A g - 108, N - 14., O - 16, C l - 35.5]$

M C l_{2}

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M C l_{3}

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M C l_{4}

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M C l

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Solution

The correct option is A $M C l_{2}$
Solution:
Equivalent weight of metal = $E$
Weight of metal chloride = $0.22 g$
Weight of silver nitrate = $0.51 g$

We know:

$\frac{w e i g h t o f m e t a l c h l o r i d e}{w e i g h t o f s i l v e r n i t r a t e} = \frac{e q . w e i g h t o f m e t a l c h l o r i d e}{e q w e i g h t o f s i l v e r n i t r a t e}$

$\frac{0.22}{0.51} = \frac{E + 35.5}{170}$

On solving we get:
$E = 37.83$

Atomic weight = $\frac{6.4}{s p e c i f i c h e a t} = \frac{6.4}{0.057} = 112.28$

Valency = $\frac{a t o m i c w e i g h t}{e q w e i g h t} = \frac{112.28}{37.83} = 2.96 \approx 3$

Molecular formula of metal chloride will be $M C l_{3}$

Option B is correct answer.

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A sample of a metal chloride weighing 0.22g required 0.51 g of AgNO3 to precipitate the chloride completely. The specific heat of the metal is 0.057. Find out the molecular formula of the chloride if the symbol of the metal is ′M′.[Ag−108,N−14.,O−16,Cl−35.5]