Grade Select GradePhysical Chemistry
loweringnof vapou pressure due to 1 molal aqueous sotion at 1000c is....
1 Answer
Beenu Mathew
Hi,
P0a – Pa/P0a = mole fraction of solute
(we all know this fact)
Now
you have been given molality= 1 mole/kg
Now this thing above can be changed to 1 mole of solute in 1000g/1kg solvent(which is the definition of molality)
mole fraction of solute = 1/1+ 1000/18 (solvent is water )
= 1/56.5
NOw, phew!! we got this done
now what was it you wanted lowering of vapor pressure
ieP0a – Pa = P0a* mole fraction of solute
= 760 torr * 1/56.5
=13.4 torr
THANKS
BEST OF LUCK!!
PLEASE APPROVE!!
:D
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