Answer

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Hint: To answer this question, we must focus on the concept of normality and how to calculate it. Normality can be described as the number of gram or mole equivalents of solute present in one liter of a solution.

Complete step by step solution:

To calculate the volume of a definite solution required to prepare solution of other Normality, we use the following equation:

\[{N_1}{V_1} = {N_2}{V_2}\]

This equation is valid for any number of solutions of different concentrations that we need to mix together.

Where,

\[{N_1}\]= initial normality

\[{V_1}\]= initial volume

\[{N_2}\]= Normality of the new solution.

\[{V_2}\]= Volume of new solution.

From the question we get,

\[{N_1}\]=0.2N of HCl

\[{V_1}\]=100ml

\[{V_2}\]=200ml of water + 100ml of Hydrochloric acid = 300ml

Substituting these values in the formula we get,

\[{N_2} = \dfrac{{{N_1}{V_1}}}{{{V_2}}} = \dfrac{{0.2 \times 100}}{{300}} = 0.667N\]

Hence, the correct answer is Option (D) 0.667N.

Additional information:

Normality is sometimes used in place of molarity because often 1 mole of acid does not completely neutralize 1 mole of base. Hence, in order to have a one-to-one relationship between acids and bases, many chemists prefer to express the concentration of acids and bases in normality.

Note: The normal concentration of a solution or normality is always equal to or greater than the molar concentration or molarity of a solution. The normal concentration can be calculated by multiplying the molar concentration by the number of equivalents per mole of solute.

Complete step by step solution:

To calculate the volume of a definite solution required to prepare solution of other Normality, we use the following equation:

\[{N_1}{V_1} = {N_2}{V_2}\]

This equation is valid for any number of solutions of different concentrations that we need to mix together.

Where,

\[{N_1}\]= initial normality

\[{V_1}\]= initial volume

\[{N_2}\]= Normality of the new solution.

\[{V_2}\]= Volume of new solution.

From the question we get,

\[{N_1}\]=0.2N of HCl

\[{V_1}\]=100ml

\[{V_2}\]=200ml of water + 100ml of Hydrochloric acid = 300ml

Substituting these values in the formula we get,

\[{N_2} = \dfrac{{{N_1}{V_1}}}{{{V_2}}} = \dfrac{{0.2 \times 100}}{{300}} = 0.667N\]

Hence, the correct answer is Option (D) 0.667N.

Additional information:

Normality is sometimes used in place of molarity because often 1 mole of acid does not completely neutralize 1 mole of base. Hence, in order to have a one-to-one relationship between acids and bases, many chemists prefer to express the concentration of acids and bases in normality.

Note: The normal concentration of a solution or normality is always equal to or greater than the molar concentration or molarity of a solution. The normal concentration can be calculated by multiplying the molar concentration by the number of equivalents per mole of solute.

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