A) \[45.0\text{ }g\text{ }conc.\text{ }HN{{O}_{3}}\]
B) \[~90.0\text{ }g\text{ }conc.\text{ }HN{{O}_{3}}\]
C) \[70.0\,g\,conc.HN{{O}_{3}}\]
D) \[54.0\,g\,conc.\,HN{{O}_{3}}\]
Correct Answer: A
Solution :
Given, molarityof 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 \[\text{=}\frac{\text{weight}\,\text{of}\,\text{HN}{{\text{O}}_{\text{3}}}}{\text{mass}\,\text{of}\,\text{HN}{{\text{O}}_{\text{3}}}\text{ }\!\!\times\!\!\text{ volume}\,\text{of}\,\text{solution}\,\text{(L)}}\] \[\therefore \]Weight of \[\text{HN}{{\text{O}}_{\text{3}}}\text{=}\,\text{molarity}\,\text{ }\!\!\times\!\!\text{ }\,\text{mol}\text{.}\,\text{mass}\,\text{ }\!\!\times\!\!\text{ }\,\text{volume(L)}\] \[=2\times 63\times \frac{1}{4}g=31.5\,g\] It is the weight of 100% \[HN{{O}_{3}}.\] But the given acid is 70%\[HN{{O}_{3}}.\] \[\therefore \]lts weight\[=31.5\times \frac{100}{70}g=45g\]
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