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:
$[M = 56; C l = 35.5]$

M C l_{3}

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M C l_{6}

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M_{2} C l_{6}

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M_{2} C l_{3}

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Solution

The correct option is C $M_{2} C l_{6}$
As we know,
Molar mass = 2 x vapour density
$2 \times 162.5 = 325$ g
Mass of $C l$ in compound $= 0.655 \times 325 = 213$ g
Mass of metal $= 325 - 213 = 112$ g
So, compound is $M_{2} C l_{6}$ .

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Similar questions

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 %

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).

Q. A transition metal

M

forms a volatile chloride which has a vapour density of

94.8

. If it contains

74.75

% of chlorine the formula of the metal chloride will be:

Q. A sample of a metal chloride weighing

0.22 g

required

0.51 g

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]

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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:[M=56;Cl=35.5]

The correct option is C M2Cl6As we know,Molar mass = 2 x vapour density 2×162.5=325gMass of Cl in compound =0.655×325=213gMass of metal =325−213=112gSo, compound is M2Cl6.

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:
$[M = 56; C l = 35.5]$

The correct option is C $M_{2} C l_{6}$
As we know,
Molar mass = 2 x vapour density
$2 \times 162.5 = 325$ g
Mass of $C l$ in compound $= 0.655 \times 325 = 213$ g
Mass of metal $= 325 - 213 = 112$ g
So, compound is $M_{2} C l_{6}$ .