Search any question & find its solution
Question:
Answered & Verified by Expert
$0.126 \mathrm{g}$ of an acid is needed to completely neutralise $20 \mathrm{mL} 0.1$ (N) NaOH solution. The equivalent weight of the acid is
Options:
Solution:
1070 Upvotes
Verified Answer
The correct answer is:
63
$\because$ Number of milli equivalents of NaOH
$\left(n^{\prime}\right)=N V=0.1 \times 20=2$
$\therefore$ Number of equivalent $=0.002$
For Neutralisation Number of equivalent of base (NaOH) Number of equivalent of acid
$0.002=\frac{W}{E}=\frac{0.126}{E}$
$\therefore \quad E=\frac{0.126}{0.002}=63$
$\left(n^{\prime}\right)=N V=0.1 \times 20=2$
$\therefore$ Number of equivalent $=0.002$
For Neutralisation Number of equivalent of base (NaOH) Number of equivalent of acid
$0.002=\frac{W}{E}=\frac{0.126}{E}$
$\therefore \quad E=\frac{0.126}{0.002}=63$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.