Question

To dissolve 0.9 gm metal 100 mL of 1N HCl is used. What is the equivalent wt. of metal?
A. 7
B. 9
C. 10
D. 6

Answer 1

Hint: The equivalent weight of metal is the mass of metal needed to generate 1 g of ${H_2}$. So to find out the equivalent weight of metal first calculate the mass of HCl by using the formula of normality.

Complete step by step answer:
Given,
Mass of metal is 0.9 gm.
Volume of HCl solution is 100 mL
Normality of HCl solution is 1N.
Normality is defined as the concentration of solute in terms of gram equivalent of solute dissolved in 1 L solution.
The formula for calculating normality is shown below.
\[Normality = \dfrac{{gram.eq.of\;solute}}{{Volume\;of\;solution}}\]
The gram equivalent of solute is given by mass of solute divided by equivalent weight.
$gram.eq.weight = \dfrac{{mass}}{{equivalent\;weight}}$
Thus, the new formula of normality is given as shown below.
$Normality = \dfrac{{mass}}{{Equivalent\;weight}} \times \dfrac{{1000}}{{Volume\;in\;mL}}$
To calculate the mass, the formula of normality is rearranged as shown below.
\[mass = \dfrac{{eq.wt \times Normality \times Volume\;in\;mL}}{{1000}}\]
The equivalent weight of HCl is 1.
To calculate the mass of HCl substitute the values in the above equation.
$\Rightarrow mass = \dfrac{{1 \times 1 \times 100}}{{1000}}$
\[\Rightarrow mass = 0.1g\]
As we know that the equivalent weight of metal is defined as the mass of metal needed to generate 1 g of ${H_2}$.
$\Rightarrow$1 g of HCl gives 1 g of ${H_2}$
If 0.9 g of metal needs 0.1 g HCl
So,
$\Rightarrow$1 g HCl will need\[\dfrac{{1 \times 0.9}}{{0.1}}\]g metal.
$\Rightarrow$mass of metal which generates 1 g ${H_2}$= 9 g.
Thus, the equivalent mass of the metal is 9 g.
Therefore, the correct option is B.

Note: As you can see in the formula of normality the division of 1000 is there. It is because the volume is present in milliliters and the calculation is done in Liters therefore to convert mL into L, the volume is divided by 1000. The equivalent weight is calculated by dividing molecular weight divided by the n factor.