Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
469 views
in Chemistry by (49.6k points)

25 cm3 of a solution of sodium hydroxide required 29 cm3 of n/10 solution of oxalic acid for neutralization. Find the normality of sodium hydroxide solution and its amount dissolved in 500 cm3 of the solution.

1 Answer

+1 vote
by (49.6k points)
selected by
 
Best answer

Substituting we have 28 × 0.1 = 25 × N2 

∴ N2 = (29 x 0.1)/25  = 0.116 

∴ Normality of sodium hydroxide solution = 0.116 

We know that W = N × E for 100 ml

i.e., Mass per dm3 = Normality × eq. nlass = 0.116 × 40 g 

∴ Mass of sodium hydroxide in 500 cm3 of the solution = (0.116 x 40)/2 = 2.32 g.

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.

Categories

...