Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 0.3 g of acid is neutralised by 40 cm3 .125N NaOH .equvivalent mass of acid is

0.3 g of acid is neutralized by 40 cm3 of 0.125N NaOH .equvivalent mass of acid is

Answers (1)

\\40 \mathrm{cm}^{3}$ of $0.125 \mathrm{N}$ NaOH neutralize acid $=0.3 \mathrm{g}$ \\ $1000 \mathrm{cm}^{3}$ of $1 \mathrm{N} \mathrm{NaOH}$ will neutralize acid\\ $=\frac{0.3}{40} \times \frac{1000}{0.125} g=60 g$ \\ $1000 \mathrm{cm}^{3}$ of $1 \mathrm{N} \ \mathrm{NaOH}$ contains one gram equivalent of the acid. Hence equivalent weight of acid=60

Posted by

lovekush

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech Admission 2025 Live: JEE Mains, SRMJEEE, MET registrations underway; MHT CET syllabus updates
anam-khan

Centre issues coaching guidelines, prohibits centres from using false claims like '100% selection'
anam-khan

BTech Admissions 2025 Live: JEE Mains, SRMJEE, VITEEE dates out; AEEE, MET registrations underway

Latest Question