Question

Vapour density of a metal chloride is 77. The equivalent mass is 3. Atomic mass of metal should be

Answer 1

Hey there!

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Answer :

Given,

Equivalent weight = 3

Now, let x be the valency of metal.

Lets, the formula of metal chloride = $MCl_{x}$

Vapour density of metal chloride = 77

Molecular weight of metal = 2 × Vapour dencity = 2 × 77 = 154

∵ Molecular weight of $MCl_{x}$ = E × x + 35.5x = (3x + 35.5)

Therefore,

x = $\frac{2 × V.D}{E +35.5 }$

⇒ $\frac{2 × 77}{3 + 35.5} = \frac{154}{35.8} = 4$

→ Therefore, the valency is 4.

∴ Now, atomic weight = Equivalent weight × valency

⇒ Atomic weight = 3 × 4 = 12

Hence, the atomic weight (mass) is 12.

Answer 2

Explanation:

Valency of metal is

Equivalent wt of metal+35.5

2×vapour density

=

3+35.5

2×77

=

38.5

154

=4.

The atomic weight of metal is equal to the product of the equivalent weight of metal and valency.

It is equal to 3×4=12.