A sample of pure sodium carbonate `0.318g` is dissolved in water and litrated with `HCI` solution. A volume of `60mL` is required to reach the methly orange end point. Calculate the molarity of the acid.

A sample of pure sodium carbonate 0.318g is dissolved in water and titrated with HCl solution. A volume of 60mL is required to reach the methyl orange end point. Calculate the molarity of the acid.

A 0.252g sample of unknown organic base is dissolved in water and titrated with a 0.14 M HCl solution .After then addition of 20mL of acid ,a pH of 10.7 is recorded. The equivalence point is reached when a total of 40mL of HCl is added . What is the molar mass of the base?

40 mL 0.05 M solution of sodium sesquicarbonate dehydrate (Na_(2)CO_(3).NaHCO_(3).2H_(2)O) is titrated against 0.05 M HCl solution, x mL of acid is required to reach the phenolphthalein end point while mL of same acid were required when methyl organe indicator was used in a separate titration. Which of the following is (are) correct statements? a. y-x = 80 mL b. y+x = 160 mL c. If the titration is started with phenolphthalein indicator and methyl orange is added at the end point, 2 x mL of HCl would be required further to reach the end point d. If the same volume of same solution is titrated against 0.10 M NaOH, x//2 mL of base would be required

2.46 g of sodium hydroxide (molar mass = 40) are dissolved in water and the solution is made to 100 cm^3 in a volumetric flask. Calculate the molarity of the solution.

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.

5.3 g of sodium carbonate have been dissolved in 500 mL of the solution. What is the molarity of the solution ?

A solution is prepared by dissolving 1.0g of NaOH in water to get 250 ml of solution. Calculato its molarity.

6.3 g of oxalic acid dihydrate have been dissolved in water to obtain a 250 ml solution. How much volume of 0.1 N NaOH would be required to neutralize 10 mL of this solution ?

1.325 g of anhydrous sodium carbonate are dissolved in water and the solution made up to 250 mL. On titration, 25 mL of this solution neutralise 20 mL of a solution of sulphuric acid. How much water should be added to 450 mL of this acid solution to make it exactly N/12 ?

1g of a monobasic acid dissolved in 200g of water lowers the freezing point by 0.186^(@)C . On the other hand when 1g of the same acid is dissolved in water so as to make the solution 200mL , this solution requires 125mL of 0.1 N NaOH for complete neutralization. Calculate % dissociation of acid ? (K_(f)=1.86(K-kg)/("mol"))