Question

The molal lowering o vapour pressure for water at 373k will be

Answer 1

Molality, M = moles of solute/kg of solvent

Mole fraction, xA = nA/nA+nB, xB= nB/nA+nB

p1 = x1po1

∴ x1 = p1/po1 = 750/760 = 0.9868

x2(solute) = 1 – 0.9868 = 0.0132

molality, m = x2/x1M1 * 1000 = 0.0132 * 1000/0.9868 * 18 = 0.7503 mol kg-1

ALTERNATIVESOLUTION :

Given that :

Temperature = 273 K

boiling point of H2O = 373 K

∴ vapour pressure H2O = 76 cm

We have,

Po-­Ps/Ps = w*M/w*M

∴ molality

= w/w*M * 1000 = Po-­Ps/Ps * 1/M * 1000

= 760 -750/750 * 1/18 * 1000

= 0.741 mol/kg of solvent

Also we have,

= 1000 *1.72 *20/50 *2

= 344

van’t Hoff factor (i) = actual mol. wt./calculate mol. wt = 172/344 = 0.5

Po-­Ps/Ps = n/n +N

∴ mole fraction = Po-­Ps/Po = 760 – 750/760

= 10/760 = 0.013