# Answer to Question #192191 in General Chemistry for Kenan

Question #192191

To neutralize 100 g of HNO3 solution, 30 cm3 of 15% NaOH solution with a density of 1.15 g / cm3 was used. Calculate the mass fraction (in%) of nitric acid in the solution!

1
2021-05-17T04:20:18-0400

Let the mass of the total solution =100 gm

So mass of the "NaOH" becomes=15 gm {That is 15% of the mass of the solution}

"d_{solution}=\\dfrac{m_{solution}}{V_{solution}}"

"V_{solution}=\\dfrac{100}{1.15} =86.95ml"

"n_{NaOH}=\\dfrac{15}{40}=0.375moles"

Molarity of NaOH "=" "\\dfrac{n_{NaOH}}{V_{solution}}\\times1000=\\dfrac{0.375}{86.95}\\times1000=4.31M"

Moles of NaOH required "=4.31\\times \\dfrac{30}{1000}=0.129 moles"

Moles of "HNO_3" needed "(n_{HNO_3})" "=0.129moles"

"n_{HNO_3}=\\dfrac{m_{HNO_3}}{molarmass}"

Mass of "HNO_3=0.129\\times63=8.127gm"

Now mass fraction of "HNO_3=\\dfrac{8.127}{100}\\times100=8.127" %

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!