A sample of peanut oil weighing `1.5763 g` is added to `25 mL "of" 0.4210 M KOH`. After saponification is complete `8.46 mL "of" 0.2732 M H_(2)SO_(4)` is needed to neutralize excess `KOH`. The saponification number of peanut oil is:

A sample of peanut oil weighing 2g is added to 25mL of 0.40M KOH . After saponification is complete, 8.5mL of 0.28M H_2SO_4 is needed to nuetralize excess of KOH . The saponification number of peanut oil is : (saponification number is defined as the milligrams of KOH consumed by 1g of oil)

A sample of peanut oil weighing 2g is added to 25mL of 0.40M KOH . After saponification is complete, 8.5mL of 0.28M H_2SO_4 is needed to nuetralize excess of KOH . The saponification number of peanut oil is : (saponification number is defined as the milligrams of KOH consumed by 1g of oil)

……………. ml of 0.1 M H_(2)SO_(4) is required to neutralize 50 ml of 0.2 M NaOH Solution :

What volume of 0.108 M H_(2)SO_(4) is required to neutralize 25.0 mL of 0.145 M KOH?

Volume of 0.1M""H_(2)SO_(4) required to neutralize 30 mL of 0.2 N""NaOH is

Volume of 0.1M""H_(2)SO_(4) required to neutralize 30 mL of 0.2 N""NaOH is