Question

To dissolve 0.9 gm metal 100 ml of 1 N HCl is used. what is the equivalent wt. of metal?

• A. 7
• B. 9
• C. 10
• D. 6

Answer 1

The correct option is B 9

Given,

$0.9gms$ metal requires $100ml$ of $1N$ $HCl$

$\Rightarrow$ Volume of $HCl=100ml$

Normality of $HCl=1N$

$\Rightarrow$ Weight of $HCl=\frac{N.Eq.Wt\times V}{1000}$

$[\because$ Normality formula$]$ $=\frac{1\times 1\times 100}{1000}$

$=0.1gm$

We know, Equivalent Weight of Metal is the mass of metal required to produce $1gm$ of $H_2⟶\left(1\right)$

Also, We know $0.1gm$ of $HCl$ gives $0.1gm$ $H_2$

$\Rightarrow$ $1gm$ $HCl$ Gives $1gm$ $H_2$

Now, if $0.9gm$ metal requires $0.1gm$ $HCl$

$\Rightarrow 1gm$ $HCl$ will require $\frac{1\times 0.9}{0.1}gm$ metal

$\therefore$ Mass of metal which produces $1gm$ $H_2=9gm$

$\therefore$ From (1) Equivalent Weight of Metal= $9gm$