Fewpal
0 votes
2.5k views
in Chemistry by (52.8k points)

6.90 gm of a metal carbonate were dissolved in 60 ml of 2(N) HCl. The excess acid was neutralized by 20 ml of 1(N) NaOH. What is the equiv. alent wt. of metal ?

(1) 40 

(2) 20

(3) 19 

(4) 39

1 Answer

+1 vote
by (58.2k points)
selected by
 
Best answer

Correct Answer is: (4) 39

Equiv. of HCl taken = 60 × 2 × 10–3

Equiv. of HCl present after the reaction = 20 × 1 × 10–3

∴ Equiv. of HCl utilized = 100 × 10–3

∴ 100 × 10–3 equiv. of metal carbonate = 6.90 gm

∴ 1 equiv. of metal carbonate

= 6.69/10-1 = 69 gm

∴ equiv. wt. of metal = 69 – 30 = 39

[becuase equiv. wt. of carbonate = 30]

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...