60.0g H2SO4/100 g solution
That is 60/98 = 0.612 moles H2SO4/100 g soln.
volume = mass/density = 100 g/0.968 = 103.3 mL and M H2SO4 = 0.612/0.1033 = 5.92M but you need to go through it more accurately. I estimated molar mass H2SO4 as 98.
To convert to molality, we want moles/kg soln. We have moles.
We have 60.0g H2SO4/100 g solution. That is 60.0g H2SO4/(60.0g H2SO4+ 40.0 g H2O). So molality =
0.612 moles H2SO4/0.040 kg H2O) = ??m
To convert to mole fraction, you want total moles in the solution. I would use the original 60.0% solution.
That is 0.612 moles H2SO4
You have 100 g soln of which 60.0 is H2SO4; therefore, 40.0 must be water.
moles H2O = 40.0/18.
Then add moles to find total moles and
XH2SO4 = moles H2SO4/total moles.
XH2O = mols H2O/total moles.
An aqueous solution of H2SO4 is 60.0 % H2SO4 by mass. The density of the solution is 0.968 g/mL .
I need to find the Molality, mole fraction, and Molarity.
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