In metal oxide, metal is 33% vapour density of metal chloride is 79 th

1.	In metal oxide, metal is 33% vapour density of metal chloride is 79 then find out the atomic weight of metal.
Answer» ANSWER : 13.712 g mol^-1 Given : Vapour density (VD) of metal chloride = 79 Metal in metal oxide = 33% To Find : Atomic weight of metal = ? Solution : Let the weight of metal oxide be 100 g. So the weight of metal will be 33 g and weight of oxygen will be 77. So equivalent weight (E_w) of metal \(= {weight\:of\:metal \over weight\:of\: oxygen } × 8\) \(={ 33\over77} × 8\) = 3.428 g We know that , Valency \(= {2 ×VD\over {E}_{w}+35.5}\) ^{\( = {2×79 \over 3.428 + 35.5}\)} ^{\(= {158\over 38.928}\)} ^{= 4.05} ^{≈ 4} ^{Now ,} ^Atomicweight⁼^valency^×^E_w = 4 × 3.428 = 13.712 g mol^-1 *Hence The atomic weight of metal is 13.712 gmol^-1*

In metal oxide, metal is 33% vapour density of metal chloride is 79 then find out the atomic weight of metal.

Answer»

ANSWER : 13.712 g mol^-1

Given :

Vapour density (VD) of metal chloride = 79

Metal in metal oxide = 33%

To Find :

Atomic weight of metal = ?

Solution :

Let the weight of metal oxide be 100 g.

So the weight of metal will be 33 g and weight of oxygen will be 77.

So equivalent weight (E_w) of metal

\(= {weight\:of\:metal \over weight\:of\: oxygen } × 8\)

\(={ 33\over77} × 8\)

= 3.428 g

We know that ,

Valency \(= {2 ×VD\over {E}_{w}+35.5}\)

^{\( = {2×79 \over 3.428 + 35.5}\)}

^{\(= {158\over 38.928}\)}

^{= 4.05}

^{≈ 4}

^{Now ,}

^Atomicweight⁼^valency^×^E_w

= 4 × 3.428

= 13.712 g mol^-1

Hence The atomic weight of metal is 13.712 gmol^-1

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