Solutions

(1) On increasing temperature solubility of gases decreases.

(2) 55.5 molal solution of $\gamma $ = 55.5 mol of

P_{$\gamma $} = K_HX_{$\gamma $}

= 2 $ \times $ 10^-5 $ \times $ ${{55.5} \over {55.5 + {{1000} \over {18}}}}$

= 10^–5 K bar

= 10^–2 bar

(3) P_{$\delta $} = K_HX_{$\delta $}

= 0.5 $ \times $ ${{55.5} \over {55.5 + {{1000} \over {18}}}}$

= 250 bar

(4) With temperature, the value of K_H (Henry’s constant) increases and solubility of gas in liquid decreases.

The vapour pressure plot for liquid $\mathrm{L}\left(\mathrm{P}_{\mathrm{L}}\right)$ versus mole fraction of $\mathrm{M}\left(x_m\right)$ is represented as :



(i) Point A on plot represents vapour pressure due to pure L (its mole fraction being 1) and point $B$ represents vapour pressure due to pure $M$. Point $Z$ represents vapour pressure due to pure liquid $\mathrm{M}$ as mole fraction of liquid $\mathrm{M}$ becomes maximum.

(ii) The dotted line represents increase in the vapour of the pure liquid $M$ with increase in its concentration. Here, pure liquid $M$ follows Raoult's law.

(iii) The red coloured curve on above the dotted line shows positive deviation shown by liquid M from the Raoult's law.

(a) Since mole fraction of pure liquid $M$ is max $\left(x_m=1\right)$ at point $Z$; hence, this point represents vapour pressure due to pure liquid $M$. There is no curve corresponding to pure liquid component L; hence, whether liquid L obey' Raoult's law or not cannot be predicted.

Option (A) is incorrect.

(b) Since mole fraction of liquid L is maximum at point B, which also corresponds to the maximum vapour pressure at point $Z$; hence, point $Z$ represents the vapour pressure of pure liquid $L$.

Option (B) is correct.

(c) At point Z, the molefraction of liquid component $\mathrm{L}$ and $\mathrm{M}$ lies above the vapour pressure curve for pure L (The dotted line). Therefore, the mixture obeys positive deviation from the Raoult's law. This means intermolecular force of attraction are lower between $M$ and L; then that between M-M and L-L.

Option (D) is correct.

When intermolecular attraction between two components A and B in the mixture is same as between A and A or B and B, hence it is a case of ideal solution. When intermolecular attraction between A and B in a mixture is smaller than that between A and A or B and B, then mixture is more vaporised, bp is lowered. It is a case of positive deviation from Raoult’s law. When intermolecular attraction between A and B is higher than that between A and A or B and B, then mixture is less vaporised, bp is increased. It is a case of negative deviation.

(a) Methanol molecules (CH)₃OH are hydrogen bonded. In a mixture of CCl₄ and CH₃OH, extent of H-bonding is decreased. Mixture is more vaporised thus, positive deviation from Raoult’s law.

(b) Acetone molecules have higher intermolecular attraction due to dipole-dipole interaction. With CS₂ , this interaction is decreased thus, positive deviation.

(c) Mixture of benzene and toluene forms ideal solution.

(d) Phenol and aniline have higher interaction due to intermolecular H-bonding. Hence, negative deviation.