`0.63 g` of diabasic acid was dissolved in water. The volume of the solution was made `100 mL`. `20 mL` of this acid solution required `10 mL` of `N//5 NaOH` solution. The molecular mass of acid is:

0.63 g of dibasic acid was dissolved in water. The volume of the solution was made 100 mL. 20 mL of this acid solution required 10 mL N/5 NaOH solution. What is the normality of acid.

5 mL of acetic acid is dissolved in 20mL of water. The volume of the solution becomes.

1.26 g of a dibasic acid were dissolved in water and the solution made up to 200 mL. 20 mL of this solution were completely neutralised by 10 mL of N/5 NaOH solution. Calculate the equivalent mass and molecular mass of the acid.

7.5 g of an acid are dissolved per litre of the solution. 20 mL of this acid solution required 25 mL of N//15 NaOH solution for complete neutralization. Calculate the equivalent mass of the acid.

0.84 g of a acid (mol wt. 150) was dissolved in water and the volume was made up to 100 ml. 25 ml of this solution required 28 ml of (N/10) NaOH solution for neutralisation. The equivalent weight and basicity of the acid

0.84 g of a acid (mol wt. 150) was dissolved in water and the volume was made up to 100 ml. 25 ml of this solution required 28 ml of (N/10) NaOH solution for neutralisation. The equivalent weight and basicity of the acid

0.126 g of acid required 20 mL of 0.1 N NaOH for complete neutralisation. The equivalent mass of an acid is