Carboxylic Acid and Acid DerivativesHard
Question
What weight of Na2CO3 of 95% purity is required to neutralize 50 mL of 0.2 N acid :-
Options
A.0.457 g
B.0.557 g
C.0.53 g
D.0.6572 g
Solution
Milli equivalent of Na2CO3 = milli equivalent of acid
× 1000 = 50 × 0.2
weight =
= 0.53 g
weight of sample = 0.53 ×
= 0.557g
× 1000 = 50 × 0.2weight =
= 0.53 gweight of sample = 0.53 ×
= 0.557gCreate a free account to view solution
View Solution FreeTopic: Carboxylic Acid and Acid Derivatives·Practice all Carboxylic Acid and Acid Derivatives questions
More Carboxylic Acid and Acid Derivatives Questions
In the reaction sequence :CH3 - C ≡ C - H CH4 + (A) (B)(B) will be:...Reduction of benzoyl chloride with Pd and BaSO4 gives:...Which of the following compounds will react with NaHCO3 solution to give sodium salt and carbon dioxide -...Acetic acid exists as dimer in a benzene solution. vant′s Hoff factor was found to be 0.52. What is the degree of ...In the following reactions - In the following reactions -(i) CH3CH2ONa A(ii) CH2 = CH2 + CO + H2O BA and B respectively ...