`1.25g` of a solid dibasic acid is completely neutralised by `25mL` of `0.25` molar Ba `(OH)_2` solution. Molecular mass of the acid is :

The pH of 0.25 M Ba(OH)_2 solution is:

0.1092 g of a dibasic acid is exactly neutralized by 21cm^(3) of 0.1N NaOH. Calculate the molecular mass of the acid.

12.5 mL of a solution containing 6.0 g of a dibasic acid in 1 L was found to be neutralized by 10 mL a decinormal solution of NaOH . The molecular weight of the acid is

0.45 g of a dibasic acid is completely neutralised with 100 ml N/10 NaOH . The molecular weight of acid is

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.

Two formulal to calculate number of milli equivalkents (mlQ) Numbr of miliequivalents =("weight")/("GEW")xx1000 Numbr of milliequivalents= volume in ml xx Normality of solution 0.09 grams of dibasic acid neutralise 40 ml of N/20 NaOH solution. Molecular weight of acid is

0.45g of an acid of mol. Mass 90 was neutralised by 20mL of 0.54 N caustic potash (KOH) . The basicity of acid is :

0.075 g of a monobasic acid required 10 mL of N//12 NaOH solution for complete neutralisation. Calculate the molecular mass of the acid

100 mL solution of an acid (molar mass 82) containing 39 g acid per litre was completely neutralized by 95.0 mL of aqueous NaOH solution containing 40 g NaOH per litre. The basicity of acid is ……

0.98 g of a polybasic acid (molar mass 98) requires 30 mL of 0.5M Ba(OH)_(2) for complete neutralization. The basicity of acid is …………