Mole concept neet

Question

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

Answer 1

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 ,}

^{Atomic weight = valency × E_w} 

= 4 × 3.428 

= 13.712 g mol^-1

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