Question

3.64 gm of a metal requires 130.0 ml of 0.5 M HCl to dissolve it. What is the equivalent weight of metal?
 $ (A) 46 $
 $ (B) 65 $
 $ (C) 56 $
 $ (D) 42 $

Answer 1

Hint: To answer the question we need to understand the concept of Molarity and Normality. Molarity and Normality are both Concentration terms. Molarity is defined as the number of moles of solute dissolved in a litre of solution. Normality is defined as the number of gram equivalents of solute dissolved in a litre of solution.

Complete Step By Step Answer:
The formula of Molarity is as under-
 $ M = \dfrac{n}{V} - - - (1) $
Where $ n $ is the number of moles which is given as given mass/Molar mass
 $ V $ is the volume of solution in litres
The formula of Normality is as under-
 $ N = \dfrac{{grmEq.}}{V} - - - (2) $
Where $ grmEq. $ is the number of gram Equivalents and it is given as the number of moles/Equivalent mass.
Equivalent Mass=Molar mass/n-factor
Here n-factor is the acidity or basicity or total positive or negative charge of base, acid, and salt respectively.
 $ V $ is the volume of solution in litres
The Molarity and Normality are related as under-
 $ Normality = Molarity \times n - factor - - - (3) $
Given in the question we have-
Given mass=3.64gm
Volume=130mL/1000=0.13L
Molarity=0.5M
Putting the values in Equation (3),
 $ N = 0.5 \times 1 = 0.5 $
(Here we have $ HCl $ so the value of n-factor will be 1 since it is a monobasic acid or the value of basicity is 1)
Now put the values in Equation (2),
 $ 0.5N = \dfrac{{3.64gm}}{{Eq.Mass \times 0.13L}} $
Now rearrange the Equation to obtain the final result.
 $ \Rightarrow Eq.Mass = \dfrac{{3.64gm}}{{0.5N \times 0.13L}} $
 $ \Rightarrow Eq.Mass = 56 $
So the correct answer is option (C).

Note:
Apart from Molarity and Normality there are several other concentration terms useful in chemistry laboratories, Pharmaceutical laboratories, and everyday lives, these include Mole fraction, Molality, Mass percentage, Volume percentage, etc. These are useful in studying Redox reactions and Acid-Base reactions.