A) 18.0
B) 21.42
C) 20.0
D) 14.21
Correct Answer: B
Solution :
[b] m-moles of \[HN{{O}_{3}}\] required |
\[=250\times 1.2=300\] |
100 g solution contain 1 mole \[HN{{O}_{3}}\] |
\[\frac{100}{1.4}mL\] solution contain 1 mole \[HN{{O}_{3}}\] |
\[C{{O}_{2}}\] molarity of \[HN{{O}_{3}}\] solution |
\[=\frac{1000}{100}\times 1.40=14\] |
\[\therefore \] \[14\times V=300\] or \[V=21.42\,mL\] |
You need to login to perform this action.
You will be redirected in
3 sec