Answer
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Hint: 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 ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$.
-The formula is weighted by gram equivalent mass of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$.
-The mass of the ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$is given that is 2 gram and the gram equivalent will be calculated by the division of molar mass by n-factor i.e. $49/1$=49.
-So, the gram equivalent mass of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ will be $\dfrac{2}{49}$.
-Now, we will calculate normality by using the formula i.e. $\dfrac{\text{Equivalent }\cdot \text{ 1000}}{\text{Volume of solution}}$
-Here, the value of equivalent and volume of solution will be 2/49 and 100 respectively.
The value of normality will be:
$\dfrac{2\ \cdot \text{ 1000}}{49\text{ }\cdot \text{ 100}}\,\text{= 0}\text{.4}$
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.
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 ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$.
-The formula is weighted by gram equivalent mass of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$.
-The mass of the ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$is given that is 2 gram and the gram equivalent will be calculated by the division of molar mass by n-factor i.e. $49/1$=49.
-So, the gram equivalent mass of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ will be $\dfrac{2}{49}$.
-Now, we will calculate normality by using the formula i.e. $\dfrac{\text{Equivalent }\cdot \text{ 1000}}{\text{Volume of solution}}$
-Here, the value of equivalent and volume of solution will be 2/49 and 100 respectively.
The value of normality will be:
$\dfrac{2\ \cdot \text{ 1000}}{49\text{ }\cdot \text{ 100}}\,\text{= 0}\text{.4}$
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.
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