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Question: Normality of 2% \({{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\) solution by volume is nearly: A. 2 ...

Normality of 2% H2SO4{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}} solution by volume is nearly:
A. 2
B. 4
C. 0.2
D. 0.4

Explanation

Solution

The concentration of a solution can be measured with the help of normality. Normality is the ratio of a number of grams equivalent by volume of solution in litre. It can measure the number of ions in the precipitation reaction.

Complete answer:
-Here, 2% of sulphuric acid means that 2 gram of sulphuric acid is present or dissolved in the 100ml of solution.
-That means the volume of a solution is 100ml.
-Now, to calculate the normality firstly we have to calculate the number of gram equivalent mass of H2SO4{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}.
-The formula is weighted by gram equivalent mass of H2SO4{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}.
-The mass of the H2SO4{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}is given that is 2 gram and the gram equivalent will be calculated by the division of molar mass by n-factor i.e. 49/149/1=49.
-So, the gram equivalent mass of H2SO4{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}} will be 249\dfrac{2}{49}.
-Now, we will calculate normality by using the formula i.e. Equivalent  1000Volume of solution\dfrac{\text{Equivalent }\cdot \text{ 1000}}{\text{Volume of solution}}
-Here, the value of equivalent and volume of solution will be 2/49 and 100 respectively.
The value of normality will be:
2  100049  100= 0.4\dfrac{2\ \cdot \text{ 1000}}{49\text{ }\cdot \text{ 100}}\,\text{= 0}\text{.4}

Therefore, option C is the correct answer.

Note: Normality is used for the acid and bases mostly. The normality is also related to the molarity o.r. normality is equal to the product of n and molarity where n is equal to the number of hydrogen ions in the acid or no. of hydroxyl ions in the base or n-factor.