What weight of `Na_(2)CO_(3) of 95%` purity would be required to neutralize `45.6 mL` of `0.235 N` acid?

What weight of Na_(2)CO_(3) of 93% purity would be required to neutralise 45.6 mL of 0.235 N acid?

What weight of Na_(2)CO_(3) of 93% purity would be required to neutralise 45.6 mL of 0.235 N acid?

The concentration of solutions can be expressed in number of ways such that Normality, Molarity, Molality, Mole fractions, Strength, % by weight, % by volume and % by strength. The molarity of ionic compound is usually expressed as formality beacuse we use formula weight of ionic compound. Addition of water to a solution changes all these terms, however increase in temperature does not change molality, mole fraction and % by weight terms. The weight of Na_(2)CO_(3) sample of 95% purity required to neutralise 45.6 mL of 0.235 N acid is:

What is the weight ( in g) of Na_(2) CO_(3) molar mass = 106 ) present in 250 mL of Its 0.2 M solution ?

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 ?

What is the weight ( in g) of Na_(2) CO_(3) (molar mass = 106 ) present in 250 mL of its 0.2 M solution ?

What is the weight (in g) of Na_(2) CO_(3) (molar mass= 106) present in 250 mL of its 0.2 M solution?

Calculate the amount of 95% pure Na_(2)CO_(3) required to prepare 5 litre of 0.5 M solution.

Calculate the amount of 95% pure Na_(2)CO_(3) required to prepare 5 litre of 0.5 M solution.