Asked by GETU
H2SO4 solution containing 200.1g of H2S04 in 0.3500l of solution at 20 degree celcius has a density of 1.3294g/ml. find a,mass percent of H2SO4 b,molarity c,molality
Answers
Answered by
bobpursley
mass percent= 200.1/(200.1+.350*1.3294)= ?
Molarity= moles/volume=(200.1/98)/.350
Molality= (200.1/98)/(.35*1.3294)
You may want to calculate molecuar mass to 4 decimal places, and not use 98.
Molarity= moles/volume=(200.1/98)/.350
Molality= (200.1/98)/(.35*1.3294)
You may want to calculate molecuar mass to 4 decimal places, and not use 98.
Answered by
Anonymous
Be careful:
Molality=mole of solute/kg of solvent
Molality= (200.1/98)/(.35*1.3294)
They solution above will return moles of solute/ g of solvent
Molality= (200.1/98)/[(.35*1.3294)*(1kg/1,000g)]=moles of solute/kg of solvent
Molality=mole of solute/kg of solvent
Molality= (200.1/98)/(.35*1.3294)
They solution above will return moles of solute/ g of solvent
Molality= (200.1/98)/[(.35*1.3294)*(1kg/1,000g)]=moles of solute/kg of solvent
Answered by
Anonymous
Nevermind. He didn't change L to mL which would return the same answer as if he did so of the proper conversions. He skipped a step and did not explain why.
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