An 8.89 g sample of an aqueous solution of nitric acid contains an unknown amount of the acid. if 27.1 ml of 0.581 m potassium hydroxide is required to neutralize the nitric acid, what is the percent by mass of nitric acid in the mixture?

The percent by mass of nitric acid in the mixture is 11.1 %

Explanation:

Step 1: Data given

Mass of HNO3 = 8.89 grams

Volume of KOH = 27.1 mL = 0. 0271 L

Molarity of KOH = 0.581 M

Step 2: The balanced equation

HNO 3  +  KOH  →  KNO 3  +  H 2 O

Step 3: Calculate the moles of KOH

Moles of KOH = molarity KOH * volume

Moles KOH = 0.581 M * 0.0271 L

Moles KOH = 0.0157 moles

Step 4: Calculate moles of HNO3

For 1 mol of KOH we need 1 mol of HNO3

For 0.0157 moles of KOH we need 0.0157 moles of HNO3

Step 5: Calculate mass of HNO3

Mass KOH = moles KOH * molar mass KOH

Mass KOH = 0.0157 moles * 63.01 g/mol

Mass KOH = 0.989 grams

Step 6: Calculate mass % HNO3 in sample

mass % = (0.989 grams / 8.89 grams)*100%

mass % = 11.1 %

The percent by mass of nitric acid in the mixture is 11.1 %

1. they vary with a higher number because boiling is way hotter than melting

level 4 to level 2 : ))