A) \[\frac{N}{15}\]
B) \[\frac{N}{20}\]
C) \[\frac{N}{16}\]
D) \[\frac{N}{18}\]
Correct Answer: C
Solution :
\[100\text{ }ml\] of \[\frac{N}{10}HCl=\frac{1}{100}g.\] \[eq=0.01\,g\,\,eq\] \[300\,\,ml\] of \[\frac{N}{20}HN{{O}_{3}}=\frac{1}{20}\times \frac{1}{1000}\times 300\] \[=0.015\,g\,\,eq\] Total acid \[=0.01+0.015\] \[=0.025\,\,geq.\] \[=\frac{1}{40}\,g\,\,eq.\] Total volume \[=400\,ml\] \[\therefore \] Normality \[=\frac{{\scriptstyle{}^{1}/{}_{40}}}{400}\times 1000=\frac{1}{16}\]You need to login to perform this action.
You will be redirected in
3 sec