Solution -
This example based on vapour pressure lowering due to additon of DDT.
Vapour pressure lowering ,
P1= X1P
here,
P1 = Vapours pressure of solution after DDT addition.
P= Vapour pressure of pure diethylether = 463.57 mmHg
X1 = mole fraction of solvent.
let's see molar mass of DDT and Diethylether from its molecular formula,
#molar mass of diethylether= C4H10O
= (12.01 x4) +(10x1.007)+(16 x 1)
= 74.12 g/mol
#molar mass of DDT = C14H9Cl5
= (12.01 x 14) + (9 x 1.007 ) + (35.45 x 5)
= 354.49 g/mol
now we have to calculate mole fraction of solution by using mass and its calculated molar mass .
Amount of DDT =10.73 g,
Moles of DDT (n) = mass (g) / molar mass of DDT
=( 10.73 g /354.49 g/mol)
= 0.0303 mol
Amount of diethylether = 237.2 g
Similarly,
moles of diethylether (n) = mass (g) / molar mass of diethylether
= (237.2 g / 74.12 g/mol )
= 3.200 moles
Now we have to calculate mole fraction of diethylether ,
XC4H10O = (3.2 mol ether/ ( 3.2 mol +0.0303 mol )
= 0.99
Now we will put mole fraction value into above formula,
Pether/DDT Sol. = 0.99 x 463.57 mm Hg
= 459.22 mm Hg
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