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The mole fraction of the solute in one molal aqueous solution is :-A one molal solution contains 1 mole of solute in 1 kg (1000 g) of water.
Moles of water $= \dfrac{1000}{18} = 55.55$
Mole fraction of the solute $= \dfrac{n_{solute}}{n_{solute} + n_{water}} = \dfrac{1}{1 + 55.55}$
$= \dfrac{1}{56.55}$
$= 0.0177 \approx 0.018$ -
The vapour pressures of two liquids 'P' and 'Q' are 80 and 60 torr, respectively. The total vapour pressure of solution obtained by mixing 3 mole of P and 2 mole of Q would be :-By Raoult's law, for an ideal solution: $P_{total} = x_P P_P^\circ + x_Q P_Q^\circ$
Total moles $= 3 + 2 = 5$
$x_P = \dfrac{3}{5} = 0.6$ and $x_Q = \dfrac{2}{5} = 0.4$
$P_{total} = 0.6 \times 80 + 0.4 \times 60$
$= 48 + 24$
$= 72$ torr -
A solution of urea (mol. mass 56 g mol$^{-1}$) boils at $100.18^\circ C$ at the atmospheric pressure. If $K_f$ and $K_b$ for water are 1.86 and 0.512 K kg mol$^{-1}$ respectively, the above solution will freeze at :-Elevation of boiling point: $\Delta T_b = 100.18 - 100 = 0.18$ K
$\Delta T_b = K_b m \Rightarrow m = \dfrac{0.18}{0.512} = 0.3516$ mol kg$^{-1}$
Depression of freezing point: $\Delta T_f = K_f m = 1.86 \times 0.3516$
$\Delta T_f = 0.654$ K
Shortcut: $\dfrac{\Delta T_f}{\Delta T_b} = \dfrac{K_f}{K_b} \Rightarrow \Delta T_f = 0.18 \times \dfrac{1.86}{0.512} = 0.654$ K
Freezing point of the solution $= 0 - 0.654 = -0.654^\circ C$
The molar mass of urea is not needed for this calculation. -
A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at $20^\circ C$ are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be -Mole ratio pentane : hexane = 1 : 4, so in the liquid $x_{pentane} = \dfrac{1}{5} = 0.2$ and $x_{hexane} = \dfrac{4}{5} = 0.8$.
Partial pressure of pentane: $p_{pentane} = 0.2 \times 440 = 88$ mm Hg
Partial pressure of hexane: $p_{hexane} = 0.8 \times 120 = 96$ mm Hg
Total vapour pressure $= 88 + 96 = 184$ mm Hg
By Dalton's law, the mole fraction of pentane in the vapour $= \dfrac{p_{pentane}}{P_{total}} = \dfrac{88}{184}$
$= 0.478$
The vapour is richer in pentane (0.478) than the liquid (0.200), because pentane is the more volatile component.