The number of grams of a dibasic acid (molecular weight \(200\)) present is \(100\, \ mL\) of its aqueous solution to give decinormal strength is:

1 g

The correct answer is option 1. 1 g.

Given,

Normality of dibasic acid \(= 0.1\, \ N\)

n-factor of dibasic acid \(= 2\)

Molar mass of dibasic acid \( = 200\, \ g\)

Volume \(= 100\, \ mL\)

\(\text{Number of moles = Molarity} × \text{Volume (in litres)}\)

\(\text{Number of moles = }0.05 × 0.1 = 0.005\)

\(\text{Number of moles} = \frac{\text{Given weight}}{\text{Molecular weight}}\)

After substituting the values we get

\(0.005 = \frac{\text{Given weight}}{200}\)

\(⇒ \text{Given weight} = 0.005 × 200\)

\(⇒ \text{Given weight} =1\, \ g\)

Therefore, 1 gram of dibasic acid should be present in 100 mL of the aqueous solution to prepare a \(0.1 N \)solution.