`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 :

9.8g of an acid of molecuar weight 98 was neutralised by 200ml of one normal caustic soda. What is the basicity of the acid?

If 20 mL of 0.4 N NaOH solution completely neutralised 40 mL of a dibasic acid, the molarity of the acid solution is

0.84 g, of an acid of Molecular weight 225 is present in 100ml of the solution. 25ml of this solution required 28 ml of N/10 NaOH solution for complete neutralisation. The basicity of the acid is

0.84 g. of an acid of Molecular weight 225 is present in 100ml of the solution. 25ml of this solution required 28ml of N/10 NaOH solution for complete neutralisation. The basicity of the acid is

0.16 gm of organic acid required 25 ml of (N)/(10) NaOH for complete neutralization. (M. wt of acids is 128). Then the basicity of acid is

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

In Kjeldahl's method 0.12 g of organic substance liberated ammonia. It is absorbed in 40ml of 0.05N acid. Which is neutralised by 27.5 ml of 0.05 N base. Calculate the percentage of nitrogen in the given compound.