A sample of nitric acid has a density of 1.41 | vedclass

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(A) The density of the solution is $d = 1.41 \ g \ mL^{-1}$.Mass percentage of $HNO_3 = 69 \%$,which means $69 \ g$ of $HNO_3$ is present in $100 \ g$

Vedclass · Accepted Answer

$15.44$

Vedclass · Answer

(A) The density of the solution is $d = 1.41 \ g \ mL^{-1}$.Mass percentage of $HNO_3 = 69 \%$,which means $69 \ g$ of $HNO_3$ is present in $100 \ g$ of the solution.Volume of $100 \ g$ of solution $= \frac{	ext{mass}}{	ext{density}} = \frac{100 \ g}{1.41 \ g \ mL^{-1}} = 70.92 \ mL = 0.07092 \ L$.Molar mass of $HNO_3 = 1.008 + 14.007 + 3 	imes 16.00 = 63.015 \ g \ mol^{-1}$.Number of moles of $HNO_3 = \frac{	ext{mass}}{	ext{molar mass}} = \frac{69 \ g}{63.015 \ g \ mol^{-1}} = 1.095 \ mol$.Molarity $(M) = \frac{	ext{moles of solute}}{	ext{volume of solution in liters}} = \frac{1.095 \ mol}{0.07092 \ L} \approx 15.44 \ M$.

$A$ sample of nitric acid has a density of $1.41 \ g \ mL^{-1}$ and the mass percentage of nitric acid is $69 \%$. Calculate the concentration of nitric acid in the sample in moles per liter. (in $M$)

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