Question

The vapor pressure of water at 25.0°C 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 22.8 torr.

Answer 1

According to Raoult's law the relative lowering of vapour pressure of a solution made by dissolving non volatile solute is equal to the mole fraction of the non volatile solute dissolved.

the relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent

$\frac{p^{0-}p}{p^{0}}=x_{B}$

Where

xB = mole fraction of solute=?

$p^{0}=23.8torr$

p = 22.8 torr

$x_{B}=\frac{23.8-22.8}{23.8}=0.042$

mole fraction is ratio of moles of solute and total moles of solute and solvent

moles of solvent = mass / molar mass = 500 /18 = 27.78 moles

putting the values

$molefraction=\frac{molessolute}{molesolute+molessolvent}$

$0.042=\frac{molessolute}{27.78+molessolute}$

$1.167+0.042(molesolute)=molessolute$

$molessolute=1.218$

mass of glucose = moles X molar mass = 1.218 X 180 = 219.24 grams