A) 45.0 g conc. \[HN{{O}_{3}}\]
B) 90.0 g conc. \[HN{{O}_{3}}\]
C) 70.0 g cone. \[HN{{O}_{3}}\]
D) 54.0 g conc. \[HN{{O}_{3}}\]
Correct Answer: C
Solution :
| Given, morality of solution = 2 |
| Volume of solution\[=250\,mL=\frac{250}{1000}=\frac{1}{4}L\]Molar mass of |
| \[HN{{O}_{3}}=1+14+3\times 16=63\,g\,mo{{l}^{-1}}\] |
| \[\because \] Molarity |
| \[=\frac{\text{weight}\,\text{of}\,\,HN{{O}_{3}}}{\text{mass}\,\text{of}\,HN{{O}_{3}}\times \text{volume}\,\text{of}\,\text{solution}\,(L)}\] |
| \[\therefore \]Weight of \[HN{{O}_{3}}=\] morality \[\times \]mol. Mass |
| \[=2\times 63\times \frac{1}{4}g=31.5g\] |
| It is the weight of \[100%\,HN{{O}_{3}}\]. |
| But the given acid is\[70%\,\,HN{{O}_{3}}\]. |
| \[\therefore \]Its weight \[=31.5\times \frac{100}{70}g=45g\] |
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