Question

32. You have 5.62 moles of HNO3. Answer the following questions based on this: (20 points) a. How many grams of HNO3 would be present with 5.62 moles of HNO3 b. What would be the molarity of HNO3 if dissolved in 560 mL of water? c. After dissolving in part b, if you diluted the solution to 1200 mL what would the molarity be? d. Would HNO3 act as a strong base, strong acid, weak base, or weak acid?

Answer 1

32.

a)

Mole = ( mass/molar mass)

Or, Mass = molar mass × moles.

Molar mass of HNO3 = 63 g/mol.

Moles of HNO3 given = 5.62.

Then, mass of HNO3

= 5.62 mol × 63 g/mol

= 354.1 g.

b)

Volume of solution = (560/1000) = 0.56 L.

Now, molarity

= ( Moles of HNO3/ volume)

= ( 5.62/0.56)

= 10.03 M.

c)

After dilution final volume

= 1200 mL = (1200/1000) = 1.2 L.

Then, final molarity = ( moles/volume)

= ( 5.62/1.2)

= 4.68 M.

d)

HNO3 incompletely ionizes in aqueous solution, hence  it is strong acid.