Need help with an Easy chemistry molality problem

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A sulfuric acid solution containing 645.6 g of H2SO4 per liter of solution has a density of 1.329 g/cm3.

(a) Calculate the mass percentage.

(b) Calculate the mole fraction.

(c) Calculate the molality.

(d) Calculate the molarity of H2SO4 in this solution.
Dec 10th, 2015

Thank you for the opportunity to help you with your question!

We have the density of the solution =1.329 g/cm^3

Consider 1000ml of the solution (1Liter)

Mass of H2SO4=645.6 g

Mass of solution =density*volume =1000*1.329 g/ml =1329 g

Mass of the water=1329-645.6=683.5

a)Mass(%)=gram of solute/gram of solution *100 =(645.6/1329) *100 =48.58%(ans)

b)Mole fraction=>

Mole of water =weight/Molecular weight =683.5/18 =37.97 mole

Mole of H2SO4=weight/Molecular weight =645.6/98=6.587

Total mole =6.587+37.97 =44.56 mole

Mole fraction =6.587/44.56 =.15 (ans)

c)Molality =moles of the solute(H2SO4)/Mass of the solvent(kg) = 6.587/.6835 =9.64 molal

d)Molarity =moles of (H2SO4)/Volume of soln =6.587/1 =6.587

I hope this would help you,but if you have any doubt regarding this,you can surely ask. :)
Dec 10th, 2015

Dec 10th, 2015
Dec 10th, 2015
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