A metal M forms a volatile chloride contaning 65.5 % chlorine. If the density of the chloride relative to hydrogen is 162.5, find the molecular formula fo the chloride (M- 56).

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Solution

The emperical formula is thus $M C l_{3}$
Emperical formula weight = 56 + $35.5 \times 3$ = 162.5
Molecular weight = $2 \times V . D$ = $2 \times 162.5$ = 325
n = $\frac{m o l e c u l a r w e i g h t}{e m p e r i c a l w e i g h t}$ = $\frac{325}{162.5}$ = 2
Molecular formula = n $\times$ emperical formula = 2 $\times$ $M C l_{3}$ = $M_{2} C l_{6}$

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A metal M, forms a volatile chloride containing 65.5% chlorine. If the density of the chloride relative to hydrogen is 162.5, find the molecular formula of the chloride. [M = 56, Cl = 35.5]

A metal M forms a volatile chloride containing 65.5% chlorine. If the density of the chloride relative to hydrogen is 162.5, find the molecular formula of the chloride.

[M = 56, Cl = 35.5]

Given: Cl = 65.5%

M = (100-65.5)% = 34.5 %

Q. A metal M forms a volatile chloride containing 65.5% chlorine. If the density of the chloride relative to hydrogen (i.e., V.D.) is 162.5, the molecular formula of the chloride is: