Question

If the mole fraction of nitric acid (HNO3) in an aqueous solution is 0.275, what is the percent by mass of HNO3?

Answer 1

Hello!

To calculate the percent by mass of HNO₃ we have to assume that we have 1 mol of solution to simplify the calculations and in that way know more easily how many moles of HNO₃ and water are in the mixture. 

If there's 1 mol of solution, then there are 0,275 moles of HNO₃ and 0,725 moles of H₂O. Now we calculate the grams of each one using the molar mas:

$gHNO_3=0,275 moles HNO_3* \frac{63,01 g HNO_3}{1 mol HNO_3}=17,3278 g HNO_3 \\ \\ gH_2O=0,725 moles H_2O* \frac{18 g H_2O}{1 mol H_2O}=13,05 g H_2O$

To finish, we calculate the percent in mass in the following way:

$=\% mass= \frac{g HNO_3}{g Total} *100= \frac{17,3278 g}{17,3278 g + 13,05 g}*100=57,04\%$

So, the percentage in mass of HNO₃ is 57,04%

Have a nice day!

Answer 2

Answer is: the percent by mass of nitric acid is 57.04 %.

mole fraction of nitric acid = 0.275.

mole fraction of nitric acid = n(HNO₃) ÷ (n(HNO₃) + n(H₂O)).

If we use 100 grams of solution:

mr = 100 g.

mr = m(HNO₃) + m(H₂O).

m(HNO₃) = 100 g - m(H₂O).

n = m/M.

mole fraction = m(HNO₃)/M(HNO₃) ÷ (m(HNO₃)/M(HNO₃) + m(H₂O)/M(H₂O)).

mole fraction  = (100 g - m(H₂O))/M(HNO₃) ÷ ((100 g - m(H₂O))/M(HNO₃) + m(H₂O)/M(H₂O)).

0.275 = (100 g - m(H₂O))/63 g/mol) ÷ ((100 g - m(H₂O))/63 g/mol + m(H₂O)/18 g/mol).

For simplier calculation, we can take m(H₂O) = x and replace to previous equatation:

(100 - x)/63 = 0.275 · ((100 - x)/63 + x/18) /×63.

100 - x = 17.325 · ((100 - x)/63 + x/18).

100 - x = 17.325 · ((100 - x + 3.5 x)/63).

100 - x = 27.5 + 0.6875x.

x = m(H₂O) = 42.96 g.

m(HNO₃) = 100 g - 42.96 g = 57.04 g.

ω(HNO₃) = 57.04 g ÷ 100 g · 100% = 57.04%.