Question

$1.12$ litre dry chlorine gas at $STP$ was passed over a heated metal when $5.56g$ of chloride of the metal was formed. What is the equivalent weight of the metal?

Answer 1

$1.12$ litre dry chloride gas at STP

$1.12$ litre dry chloride gas $=$ $1$ mole of gas occupies $22.4$ litre.

$1.12$ litre gas $=$ ${}^{\prime }{x}^{\prime }$ moles.

$22.4$ litre $=$ $1$ moles.

by cross multiplication

${}^{\prime }{x}^{\prime }=\frac{1.12}{22.4}$

$1.12$ litre gas has $0.05$ moles of dry chloride gas mass of $0.05$ moles of dry chloride gas.

mass of $1$ mole chloride $=35.5$ gm.

mass of $0.05$ mole chloride $=4$ gm

by cross multiplication$y=35.5\times 0.05$

$=1.775$ gm

$1.12$ litre dry chloride gas has $1.775$ gm chloride.

Given that $5.56$ gm of metal chloride is obtained.

That means $5.56$ gm of metal chloride contains $1.775$ gm of chloride.

Therefore, the moss of metal in the metal chloride is $5.56-1.775=3.755$ gm

$5.56$ gm of metal chloride contains $3.785$ gm metal and $1.775$ gm chloride

Equivalent weight of the metal $=\frac{massofmetal}{massofchloride}\times molarmassofchloride$

$=\frac{3.785}{1.775}\times 35.5=75.27$

Equivalent weight of metal $=75.27$ gm