Question

One mole of chlorine combines with certain weight of a metal giving 111g of its chloride. The same amount of metal can displace 2g hydrogen from an acid. The atomic weight of the metal is:

• A. $40$
• B. $20$
• C. $80$
• D. None of these

Answer 1

The correct option is A $40$

$M^{+2}+C{l}_2\to MgC{l}_2$
mole =$\frac{\omega }{M.\omega t}=\frac{111}{(M.\omega t{)}_{\omega }+2(M.\omega t{)}_{CI}}$
mole should be some
$\frac{\omega }{M.\omega t}=\frac{111}{M.\omega t+71}--\left(1\right)$
Now metal also can displace 29 Hydrogen
so mole should be same
so $\frac{\omega }{M.\omega t}=\frac{2}{2}=1$
put $\frac{\omega }{(M.\omega t)}$ in eq (1)
$1=\frac{111}{M.\omega t+71}$
$M.\omega t=111-71$
$=40$