Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

$HNO_3$ used as a reagent has specific gravity of $1.42 mL^{-1}$ and contains $70\%$ by strength $HNO_3$. Calculate Normality of acid


Can you answer this question?

1 Answer

0 votes
Given, Strength of $HNO_3 = 70\%$
Volume of the solution = 100mL
Volume of $HNO_3$ = 70 mL
$\therefore$ Weight of $HNO_3$ in the solution = $70\times1.42g$
$\therefore$ Equivalent weight of $HNO_3$ in the solution = $\large\frac{70\times1.42}{63}$
Since N=$\large\frac{\text{Equivalent weight of $HNO_3$ in the solution}}{\text{V in litre}}$
$N_{HNO_3} = \large\frac{70\times1.42}{63\times\large\frac{100}{1000}} = 15.78$
Hence answer is (a)
answered Feb 11, 2014 by sharmaaparna1
edited Mar 18, 2014 by mosymeow_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App