The vapor pressure of water at 25.0ºC is 23.8 torr. Determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.0 g of water to change the vapor pressure to 23.3 torr.
Solution-
First we will find the mole fraction of water (n):
=>P(solu) = n Pv(water)
Pv(solu) = 23.3 and Pv(water) = 23.8 so by substitution:
=>23.3 = n * 23.8
n= 0.978
Now the moles of glucose:
Now the moles of water = 500 g / 18 (molar weight)= 27.7 mol
=>n = moles of water / ( moles of water + moles of
glucose)
=> 0.978 = 27.7 / ( 27.7+ moles of
glucose)
=>0.978 moles of glucose + 27.09 = 27.7
=>0.978 moles of glucose = 0.61
moles of glucose = 0.62 mol
Now the mass of glucose = noumber of glucose moles * molar
mass
=
0.62 * 180
= 111.6 g
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