Need help with an Easy chemistry molality problem

timer Asked: Dec 10th, 2015
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Question Description

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.

Tutor Answer

School: Carnegie Mellon University

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. :)

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