Question

To prepare one liter 0.05 N $ HCl $ solution from 0.5 N $ HCl $ solution, volume of water should be added is equal to $ x\text{ x 1}{{\text{0}}^{2}}\text{ ml} $ , value of ‘x’ is:

Answer 1

Hint The solutions are 0.5 N and 0.05 N, and the solution is being converted from the higher concentration to the lower concentration. N means the normality of the solution, and it is calculated by dividing the gram equivalent of the solute by the volume of the solution in liters only.

Complete step by step answer:
There are many factors that can define the concentration of the solution like molarity, molality, mole fraction, ppm, etc, and normality is one of the major factors. The two factors by which we can calculate the normality of the solution are the gram equivalent of the solute and the volume of the solution. And the volume of the solution should be taken in liters, so the normality is calculated by dividing the gram equivalent of the solute by the volume of the solution.
The solutions are 0.5 N and 0.05 N, and the solution is being converted from a higher concentration to a lower concentration.
So, the 0.5 N solution is changing into 0.05 N solution, therefore, the normality is increasing by $ \dfrac{1}{10}\text{ times} $ . So, we can write:
 $ Normality=\dfrac{1}{10} $
By the formula of normality, we can write:
 $ Normality=\dfrac{Gram\text{ }equivalence}{Volume\text{ }of\text{ }the\text{ }solution} $
Combining, the above we can write:
 $ Normality=\dfrac{Gram\text{ }equivalence}{Volume\text{ }of\text{ }the\text{ }solution}\text{ x }\dfrac{1}{10} $
Suppose, the initial solution was taken as 1 Liter and we can write this as 1000 ml, we get:
 $ Normality=\dfrac{Gram\text{ }equivalence}{1000}\text{ x }\dfrac{1}{10} $
 $ Normality=\dfrac{Gram\text{ }equivalence}{10000}\text{ } $
This volume has to be written in the form of $ x\text{ x 1}{{\text{0}}^{2}}\text{ ml} $ ,
So, it will be $ \text{100 x 1}{{\text{0}}^{2}}\text{ ml} $ , therefore, the value of 100.

Note: Since there is the involvement of volume in the calculation of the normality of the solution, so the normality is temperature-dependent and changes even on changing the temperature of the solution.