Question

11. How many grams of concentrated nitric acid solution
should be used to prepare 250 mL of 2 M HNO?
The concentrated acid is 70% (w/w) HNO3-
[NEET-2013]
(1) 90.0 g conc. HNO, (2) 70.0 g conc. HNO,
(3) 54.0 g conc. HNO3 (4) 45.0 g conc. HNO3​

Answer 1

$\huge{\mathfrak{\red{\underline{\underline{Answer:-}}}}}$

$\sf{45\;g\;of\;HNO_{3}\;is\;required.}$

$\huge{\mathfrak{\red{\underline{\underline{Explanation:-}}}}}$

$\sf{To\;prepare\;250ml\;of\;2M\;HNO_{3}\;from\;70\%\;concentrated\;nitric\;acid:}$

$\sf{Morality\;of\;HNO_{3}=\dfrac{Moles\;of\;HNO_{3}}{Volume\;of\;solution}\times 1000}$

$\sf{2=\dfrac{Moles\;of\;HNO_{3}}{250}\times 1000}$

$\sf{Moles\;of\;HNO_{3}=0.5\;mol}$

$\sf{Mass\;of\;0.5\;mol\;of\;HNO_{3}=Molar\;mass\;of\;HNO_{3}\times Moles\;of\;HNO_{3}}$

$\sf{=63\times 0.5=31.5\;g}$

$\sf{70\%\;concentrated\;nitric\;acid\;means,\;70g\;of\;HNO_{3}\;is}$$\sf{present\;in\;100g\;of\;HNO_{3}\;solution.}$

$\sf{Therefore\;31.5g\;of\;HNO_{3}\;is\;present\;in\;\;\dfrac{100\times 31.5}{70}=45g}$

${\boxed{\boxed{\bf{\red{Hence,\;45g\;of\;HNO_{3}\;required.}}}}}$

Answer 2

${\mathfrak{\blue{\underline{\underline{Answer:-}}}}}$

$\sf{Option\;(4)\;is\;correct\;45g\;conc.\;HNO_{3}}$

${\mathfrak{\blue{\underline{\underline{Explanation:-}}}}}$

${\tt{\underline{Given:-}}}$

$\sf{Molarity\;of\;solution=2M}$

$\sf{Volume\;of\;solution=250\;ML=\dfrac{250}{1000}=\dfrac{1}{4}L}$

$\sf{Molar\;mass\;of\;HNO_{3}=1+14+3\times16=63g\;mol^{-1}}$

$\sf{Molarity=\dfrac{Weight\;of\;HNO_{3}}{Mass\;of\;HNO_{3}\times Volume\;of\;solution}}$

$\sf{Therefore,}$

$\sf{Weight\;of\;HNO_{3}=Molarity\times mol.\;mass\times volume}$

$\sf{=2\times 63\times\dfrac{1}{4}=31.5\;g}$

$\sf{It\;is\;the\;weight\;of\;100\%\;HNO_{3}}$

$\sf{But\;the\;given\;acid\;is\;70\%\;HNO_{3}}$

$\sf{Therefore,}$

$\sf{It's\;Weight=31.5\times \dfrac{100}{70} g=45g}$

${\boxed{\boxed{\bf{\red{Solved}}}}}$