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Can you please help me with this problem and show me how to do it?
The vapor pressure of pure water at 40 degrees Celsius is 55.3 torr. Calculate the mass of propylene glycol (C3H8O2) that must be added to 0.340 kg of water to reduce the vapor pressure by 2.88 torr at 40 degrees Celsius.
1 個解答
- Dr WLv 75 年前
via raoults law
.. Pa = P*a x χa
where
.. Pa = partial pressure of component "a" over the liquid
.. P*a = vapor pressure of pure component "a"
.. χa = mole fraction of component "a"
method
.. (1) assume PEG.. (propylene glycol).. is non-volatile.. with vapor pressure = 0
.. .. ..so that all the pressure is due to H2O only.
.. (2) use that equation to determine mole fraction of water
.. (3) convert 0.340kg to mole water
.. (4) use the equation.... mole fraction H2O = mole H2O / (mole H2O + mole PEG)
.. .. ...to calculate mole PEG
.. (5) convert moles PEG to mass PEG
YOU try it.
********
more hints.
.. Pa = 55.3torr - 2.88torr = __ torr
.. P*a = 55.3torr
.. χa = Pa / P*a = __ torr / __ torr = ___... ... raoults law
even more hints
.. 0.340kg H2O x (1000g / 1kg) x (1mol / 18.02g) = ___mol
then
.. (__ mol H2O) / (__ mol H2O + Z mole PEG) = ___ (χa from above)
fill in the blanks then solve for Z then
.. __ mol PEG x (__g PEG / mol PEG) = ___g PEG
