How many grams of concentrated nitric acid so | vedclass

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(C) Step $1$: Calculate the number of moles of $HNO_3$ required.$n = M 	imes V(L) = 2.0 \ mol/L 	imes 0.250 \ L = 0.5 \ mol$.Step $2$: Calculate the

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$45.0 \ g$ conc. $HNO_3$

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(C) Step $1$: Calculate the number of moles of $HNO_3$ required.$n = M 	imes V(L) = 2.0 \ mol/L 	imes 0.250 \ L = 0.5 \ mol$.Step $2$: Calculate the mass of pure $HNO_3$ required.Molar mass of $HNO_3 = 1 + 14 + (3 	imes 16) = 63 \ g/mol$.Mass of $HNO_3 = 0.5 \ mol 	imes 63 \ g/mol = 31.5 \ g$.Step $3$: Calculate the mass of the $70\%$ concentrated solution.Since the solution is $70\%$ $HNO_3$ by mass,$70 \ g$ of $HNO_3$ is present in $100 \ g$ of solution.Mass of solution $= (100 \ g 	ext{ solution} / 70 \ g 	ext{ } HNO_3) 	imes 31.5 \ g 	ext{ } HNO_3 = 45 \ g$.

How many grams of concentrated nitric acid solution should be used to prepare $250 \ mL$ of $2.0 \ M \ HNO_3$? The concentrated acid is $70\% \ HNO_3$ by mass.

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