for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.
pA ∝ xA
Where only solvent is volatile
pA a xA where p A is vapour pressure of solvent having mole fractionxA,
But xA + xB = 1
xA = 1 – xB where xB is mole fraction of non-volatile solute B
pA = p0A (1 – xB)
= p0A – p0A x B
Total vapour pressure
Solution containing non-volatile solute : For a solution of non-volatile solid in a liquid the vapour pressure contribution by the non-volatile solute is negligible. Therefore the partial vapour pressure of a solution containing a non-volatile solute is equal to the product of the vapour pressure of the pure liquid (solvent p0A) and its mole fraction in solution.
PA = P0A x xB ....(i)
xB is the mole fraction of the non-volatile solute
B, then xA + xB = 1
xA = 1 – xB ....(ii)
Substituting the value of xA fromeq. (ii) into eq. (i), we get, pA = p0A (1 – xB) = p0A – p0A x B
Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.
(a) Mol. mass of
Volume of solution = 4.3 L
(b) Number of moles present in 1000 ml of 0.5M H2SO4= 0.5 mol
therefore number of moles present in 30ml of 0.5M H2SO4=mol =0.015mol
therefore molarity =0.015/0.5L
thus molarity is 0.03M
Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.
Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as: