A) 45.0 g cone. \[\text{HN}{{\text{O}}_{\text{3}}}\]
B) 90.0 g conc.\[\text{HN}{{\text{O}}_{\text{3}}}\]
C) 70.0 g cone. \[\text{HN}{{\text{O}}_{\text{3}}}\]
D) 54.0 g conc.\[\text{HN}{{\text{O}}_{\text{3}}}\]
Correct Answer: A
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 \[=\frac{weight\,of\,HN{{O}_{3}}}{mass\,of\,HN{{O}_{3}}\times volume\,of\,solution\,(L)}\] \[\therefore \]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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