Question

$71g$ chlorine combines with certain weight of a metal giving $111g$ of its chloride metal is

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

Answer 1

The correct option is A $40$

Let metal be ${}^{\prime }{x}^{\prime }$ and there molecular mass be ${}^{\prime }{y}^{\prime }$.

$1C{l}_2{+}^{\prime }{x}^{\prime }=nxC{l}_2$

$n=$ no. of mole

$[1mole=71gm]$

$[1chlorineatom=35.5gm]$

See that as it replace $2gH=$ replacing $2$ moles of $H$ or $2^{+}$ charge.

thus metal has $2+$ charge

$\therefore$ type of metal chloride salt is $xC{l}_2$

$\therefore$ We know that

$C{l}_2+y=xC{l}_2$ $y=$ molecular mass for $1mole$

$71+y=111$

$y=111-71$

$y=40$

$\therefore$ Chloride metal is $40gm$.