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# What amount of water is added in 40 ml of NaOH (0.1N) which is neutralized by 50ml of HCl (0.2N)?(a) 80 ml(b) 60 ml(C) 40 ml(d) 90 ml  Hint: The above numerical can be solved on the concept of normality dilution. Dilution is basically the process of preparing a less concentrated solution from the more concentrated one.

Complete step by step solution:
> The concentration of a solution could also be expressed in terms of Normality. It is based on an alternate chemical unit of mass called the equivalent weight. The normality of a solution is the concentration expressed as the number of equivalent weights (equivalent) of solute per liter of solution. In a chemical mixture 1 normal (1 N) solution contains 1 equivalent weight of solute per liter of solution. Since, normality simplifies the calculation required for chemical concentration; it is widely used in analytical chemistry.
> It is based on the concept of dilution. Use the relationship that the number of gram equivalent of moles before dilution is equal to the number of gram equivalent of moles after dilution.
${{\text{N}}_{1}}{{\text{V}}_{1}}={{\text{N}}_{2}}{{\text{V}}_{2}}$
Let the volume of NaOH be x.
(NaOH) ${{\text{N}}_{1}}{{\text{V}}_{1}}={{\text{N}}_{2}}{{\text{V}}_{2}}$ (HCl)
$\Rightarrow 0.1\text{x }x=0.2\text{x50}$
$\Rightarrow x=\dfrac{0.2\text{x50}}{0.1}=100$
Therefore, amount of NaOH is 100ml
Since, 40ml of NaOH is there, so, amount of water added=(100-40)ml = 60ml
Therefore, the correct option is (b) 60ml.

Note: It should be noted that the number of parts by weight of a substance (element or compound) that will combine with or displace, directly or indirectly 1.008 parts by weight of hydrogen, 8 parts by weight of oxygen and 35.5 parts by weight of chlorine is known as equivalent weight.
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