**How to calculate the requirement of commercially available Concentrated Hydrochloric Acid to prepare 0.1N or 0.1M HCl Solution?**

I am asking this question to many fresher candidates in interview; to evaluate their understanding of Chemistry, Maths, Convincing power etc.

To calculate the required quantity of commercially available Concentrated Hydrochloric Acid to prepare 0.1N or 0.1M HCl solution following steps are to be understood,

Concentration of collmericially available concentrated Hydrochloric Acid (HCl) = 32% v/v to 37% v/v (__Why?__)

Let us consider 35% v/v (for example).

Let us take an example; we want to prepare 500 ml of 0.1N or 0.1M HCl.

__Step 1 Calculate the Mass of HCl in 1000 ml 35% v/v Concentrated Hydrochloric Acid (HCl):__

35% v/v Concentrated Hydrochloric Acid (HCl) = 36 ml HCl in 100 ml Water

So, 360 ml HCl in 1000 ml Water

The Density of HCl = 1.1789 g/ml

Thus; Mass of HCl in 1000 ml water = 360 ml (HCl) x 1.1789 g/ml Density of HCl

**= 424.404 g**

__Step 2 Calculate Molecular Weight of HCl:__

Molecular Weight of HCl = 1.0078 g/mole (H) + 35.453 g/mole (Cl)

** = 36.4608 g/mole**

__Step 3 Calculate Molarity of 35% v/v Concentrated Hydrochloric Acid (HCl)__

Moles of HCl in 1000 ml water = Mass of HCl in 1000 ml water / Molecular Weight of HCl

= 424.404 g / 36.4608 g/mole

**= 11.6400 Moles **

Hence, Molarity of 35% v/v Concentrated Hydrochloric Acid (HCl) = **11.64 M**

__Step 4 Calculate The requirement of commercially available Concentrated Hydrochloric Acid to prepare 0.1N or 0.1M HCl Solution:__

Now,

We will use the Dilution equation i.e. M1 x V1 = M2 x V2

M1 = 11.64 M [Molarity of 35% v/v Concentrated Hydrochloric Acid (HCl)]

V1= We have to find out

M2= 0.1 M [The concentration of HCl which we want to prepare]

V2= 500 ml [The Volume of HCl which we want to prepare]

So; V1 = M2 x V2 / M1

=0.1 M x 500 ml / 11.64 M

= **4.2955 ml (**Approximately** 4.3 ml)**

__The answer is __

__The requirement of commercially available Concentrated Hydrochloric Acid to prepare 0.1N or 0.1M HCl Solution is 4.30 ml __

Note: The Acidity of HCl is one hence, the Normality (N) and Molarity (M) both are same.

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