Question

How many grams of dibasic acid (mol. Wt. 200 g) should be present in 100 mL of the aqueous solution to give 0.1 N?
${\text{A}}{\text{.}}$ 1 g
${\text{B}}{\text{.}}$ 2 g
${\text{C}}{\text{.}}$ 3 g
${\text{D}}{\text{.}}$ 4 g

Answer 1

Hint- Here, we will proceed by finding the equivalent weight of the solute (dibasic acid). Then, we will find the number of grams equivalent of the dibasic acid. Finally, we will give the general formula for normality of solution.

Complete answer:
Formulas Used- Equivalent Weight = $\dfrac{{{\text{Molecular Weight}}}}{{{\text{n - factor}}}}$, Normality of the solution (in N) = $\dfrac{{{\text{Number of gram equivalent}}}}{{{\text{Volume of the solution (in L)}}}}$ and Number of gram equivalent = $\dfrac{{{\text{Weight of solute}}}}{{{\text{Equivalent weight of solute}}}}$.
Molecular weight of dibasic acid = 200 g
Volume of the aqueous solution = 100 mL = $\dfrac{{100}}{{1000}}{\text{ L}} = \dfrac{1}{{10}}$ = 0.1 L
Normality of the aqueous solution = 0.1 N
Let the weight of dibasic acid present be W grams
i.e., Weight of dibasic acid = W g
A dibasic acid yields two free hydrogen ions in solution for each ionized acid molecule, which has two replaceable hydrogen atoms in other words.
So, n-factor of the dibasic acid (number of ${{\text{H}}^ + }$ ions present) = 2
As we know that the equivalent weight is given by the formula,
Equivalent Weight = $\dfrac{{{\text{Molecular Weight}}}}{{{\text{n - factor}}}}$
Using the above formula, we get
Equivalent weight of dibasic acid = $\dfrac{{{\text{Molecular weight of dibasic acid}}}}{{{\text{n - factor of dibasic acid}}}} = \dfrac{{200}}{2}$ = 100 g
Also we know that for any solution, the formula for normality (in N) is given by
Normality of the solution (in N) = $\dfrac{{{\text{Number of gram equivalent}}}}{{{\text{Volume of the solution (in L)}}}}{\text{ }} \to {\text{(1)}}$
Number of gram equivalent = $\dfrac{{{\text{Weight of solute}}}}{{{\text{Equivalent weight of solute}}}}{\text{ }} \to {\text{(2)}}$
Here, the solute is dibasic acid
Using the formula given by equation (2), we can write
Number of gram equivalent of dibasic acid = $\dfrac{{{\text{Weight of dibasic acid}}}}{{{\text{Equivalent weight of dibasic acid}}}}$
$ \Rightarrow $ Number of gram equivalent of dibasic acid = $\dfrac{{\text{W}}}{{100}}{\text{ }} \to {\text{(3)}}$
Using the formula given by equation (1), we can write
Normality of the aqueous solution (in N) = $\dfrac{{{\text{Number of gram equivalent of dibasic acid}}}}{{{\text{Volume of the aqueous solution (in L)}}}}$
By substituting the known values in above equation, we get
$
   \Rightarrow 0.1 = \dfrac{{\left( {\dfrac{{\text{W}}}{{100}}} \right)}}{{0.1}} \\
   \Rightarrow 0.1 = \dfrac{{\text{W}}}{{100 \times 0.1}} \\
   \Rightarrow 0.1 = \dfrac{{\text{W}}}{{10}} \\
   \Rightarrow {\text{W}} = 10 \times 0.1 \\
   \Rightarrow {\text{W}} = 1{\text{ g}} \\
 $
Therefore, the required grams of dibasic acid present in the aqueous solution is 1 g.

Hence, option A is correct.

Note- Normality is a concentration factor equal to the gram weight equivalent per litre of solution. Gram equivalent mass is a measure of a molecule's reactive capacity. The position of the solute in the reaction determines the normality of the solution. Normality is also called the concentration counterpart of a solution.