The vapour density of volatile chloride of metal is 59.58 equal at mos

1.	The vapour density of volatile chloride of metal is 59.58 equal at most of the metal is 24 the atomic mass of element will be
Answer» HEYA---- Given Vapour Density ( VD ) = 59.5 i.e. Molecular Weight = ( VD 2 ) = 119 Now, No of max Cl atoms that the compound can have is TWO because three exceeds the given molecular weight of the compound i.e. Metallic Component is = 119 - 71 = 48 also the equivalent weight of the METAL is given as 24 and there are 2 CHLORINE atoms in the compound which implies that in this case Mol.wt = 2 Eq.wt = 24*2 = 48……. So 48 is the atomic mass of the metal( which happens to be Cadmium )…… [email protected]

The vapour density of volatile chloride of metal is 59.58 equal at most of the metal is 24 the atomic mass of element will be

Answer»

HEYA----

Given Vapour Density ( VD ) = 59.5

i.e. Molecular Weight = ( VD *2 ) = 119

Now, No of max Cl atoms that the compound can have is TWO because three exceeds the given molecular weight of the compound

i.e. Metallic Component is = 119 - 71 = 48

also the equivalent weight of the METAL is given as 24 and there are 2 CHLORINE atoms in the compound which implies that in this case

Mol.wt = 2 * Eq.wt = 24*2 = 48…….

So 48 is the atomic mass of the metal( which happens to be Cadmium )……