Question

How many grams of dibasic acid (mol. wt. 200) should be present in the 100 ml of the aqueous solution to give 0.1 N?
A. 1g
B. 2gm
C. 10g
D. 20g

Answer 1

Hint: To solve the following question, we need to know about two different terms and they are: -
Dibasic acid – The term dibasic acid is commonly referred to any substance that can donate two protons or hydrogen ions per molecule in an acid base reaction.
Aqueous Solution – An aqueous solution is a solution in which the solvent is water. It is mostly shown in chemical equations by appending (aq.) to the relevant chemical formula. For e.g. A solution of table salt.

Complete step by step answer:
We are given that: -
Molecular Weight = 200 g
Weight of aqueous solution = 100 ml
Now,
Equivalent weight of dibasic acid = $\dfrac{{Mol.wt.}}{{Basicity\; of \; acid}}$
                                                           = $\dfrac{{200}}{2}$ = 100g
Because the basicity of Dibasic acid is 2.

Now, Strength-0.1 N, V= 100 ml
and we have to find the grams of Dibasic acid that is present in 100ml of solution. It is represented by ‘x’ in following equation:
Formula for finding x;

x= $\dfrac{{EVN}}{{1000}}$
Put all the values given,

= $\dfrac{{100 \times 100 \times 0.1}}{{1000}}$= 1g

Hence the correct answer is A. i.e. 1

Note: Thus, Dibasic acid is an acid containing two potential protons to donate and aqueous solution is the solution in which solvent is water.
Equivalent weight of dibasic acid = $\dfrac{{Mol.wt.}}{{Basicity of Acid}}$
andx= $\dfrac{{EVN}}{{1000}}$.