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 \[V=21.42\,mL\] |
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