Question

At 27 degrees Celcius, the vapor pressure of pure water is 23.76 mmHg and that of an aqueous solution of urea 22.95 mmHg. Calculate the molality of urea in the solution.

Answer #1

lowering of vapour pressure

(p0-p)/p0 = x

p0: vapor pressure of pure solvent = 23.76 mm Hg

p: vapor pressure of solution = 22.95 mm Hg

x2: mole fraction of solute

(23.76 - 22.95) / 23.76 = x

x = 0.034

whis much less than one

generally formula for mole fraction

x = n2 / n2 + n2 since n2 tooles we rearrange equation

x = n2/ n1

n2 = moles of solute

n1 = no of moles of solvent

i can write the above equation

mole fraction x = n2 / (weight of solvent / moolar mass of solvent)

n2 / mass of the solvent = x / ( molar mass of the solvent)

multply with thousend both sides

n2 (1000) / mass of the solvent in grams = x (1000) / molar mass of the solvent)

left side term will indicate the molality

m = 0.034 (1000) / 18

= 34 / 18

= 1.88 molal

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