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
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