How do I solve this AP Chemistry problem on H2SO4?

Question

Concentrated sulfuric acid (18.4M H2SO4) has a density of 1.84 grams per millimeter. After dilution with water to 5.20M, the solution has a density of 1.30 grams per milliliter and can be used as an electrolyte in lead storage batteries for automobiles.

 

A.Calculate the volume of concentrated acid to prepare 1.00L of 5.20M H2SO4. I calculated 0.283L

 

B. Determine the mass percent of H2SO4 in the original concentrated solution.

 

C. Calculate the volume of 5.20M H2SO4 that can be completely neutralized with 10.5 grams of sodium bicarbonate( NaHCO3)

 

D. What is the morality of the 5.20M H2SO4 

Answer 1

A.  (x ml)(18.4 M) = 1000 ml (5.20 M)

x = 283 ml = 0.283 L (you're answer is correct)

 

B.  18.4 mol/L x 98.08 g/mole = 1805 g/L of H2SO4

1000 ml x 1.84 g/ml = 1840 g of solution

1805g/1840g (x100%) = 98.1% by weight

 

C.  H2SO4 +2 NaHCO3 ==> Na2SO4 + 2H2O + 2CO2 ... balanced equation

moles NaHCO3 used = 10.5 g x 1 mole/84.0 g = 0.125 moles

moles H2SO4 neutralized = 0.125 moles NaHCO3 x 1 moles H2SO4/2 moles NaHCO3 = 0.0625 moles H2SO4

volume H2SO4 = (x L)(5.20 mol/L) = 0.0625 moles and x = 0.01209 L = 12.1 ml

 

D.  Are you asking for the MOLALITY of H2SO4?  That is moles/kg solvent. 

moles in 1 liter = 5.20 moles

mass of 1 liter = 1.30g/m x 1000 ml = 1300 g

mass of solvent = 1300 g - mass of H2SO4 and mass H2SO4 = 5.20 moles x 98.08 g/mole = 510 g

mass of solvent = 1300 g - 510 g = 790 g = 0.790 kg

molality (m) = 5.20 mole/0.790 kg = 6.58 m