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**Q.35. Suggest the most important type of intermolecular attractive interaction in the following pairs.****(i) n-hexane and n-octane ****(ii) I _{2} and CCl_{4}**

The order of increasing polarity is:

Cyclohexane < CH

Therefore, the order of increasing solubility is:

KCl < CH

Therefore, molality of Na

It can be observed from the graph that the plot for the *p*_{total} of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.**Q.****40. Calculate the mass percentage of aspirin (C _{9}H_{8}O_{4}) in acetonitrile (CH_{3}CN) when 6.5 g of C_{9}H_{8}O_{4} is dissolved in 450 g of CH_{3}CN.**

Then, total mass of the solution = (6.5 + 450) g = 456.5 g

Therefore, mass percentage of C

Percentage of nitrogen (N

Also, it is given that water is in equilibrium with air at a total pressure of 10 atm,

that is, (10 × 760) mm Hg = 7600 mm Hg

Therefore,

Partial pressure of oxygen,

= 1520 mm Hg

Partial pressure of nitrogen,

= 6004 mmHg

Now, according to Henry’s law:*p *= *K*_{H}.*x*

For oxygen:

= 4.61 x 10^{-5}

For nitrogen:

Hence, the mole fractions of oxygen and nitrogen in water are 4.61 × 10^{−5 }and 9.22 × 10^{−5} respectively.**Q.****42. Benzene and toluene form ideal solution over the entire range of composition.The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.****Ans. **Molar mass of benzene (C_{6}H_{6}) = 6 × 12 + 6 × 1 = 78 g mol^{-1}

Molar mass of toluene (C_{6}H_{5}CH_{3}) = 7 × 12 + 1 × 8 = 98 g mol^{-1}

Now, no. of moles present in 80 g of benzene

And, no. of moles present in 100 g of toluene

∴ Mole fraction of benzene,

And, mole fraction of toluene, x_{t }= 1 - 0.486 = 0.514

It is given that vapour pressure of pure benzene, P°_{b}= 50.71 mm Hg

And, vapour pressure of pure toluene, P°_{t} = 32.06 m Hg

Therefore, partial vapour pressure of benzene, p_{b }= x_{b} × p°_{t}

= 0.486 × 50.71

= 24.645 mm Hg

Hence, mole fraction of benzene in vapour phase is given by:

= 0.599

= 0.6**Q.****43. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.****Ans.** ** ** ** **

Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H^{ }ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order:

Acetic acid < trichloroacetic acid < trifluoroacetic acid**Q.****44. Calculate the depression in the freezing point of water when 10 g of CH _{3}CH_{2}CHClCOOH is added to 250 g of water. K_{a} = 1.4 × 10^{−3}, K_{f }= 1.86 K kg mol^{−1}.**

∴ No. of moles present in 10 g of CH

= 0.0816mol

It is given that 10 g of CH

∴ Molality of the solution,

= 0.3264 mol kg

Let

Since *α* is very small with respect to 1, 1 − α ≈ 1

⇒ K_{a} = Cα^{2}

= 0.0655

Again

Total moles of equilibrium = 1 − α + α + α

= 1 + α

Hence, the depression in the freezing point of water is given as:

ΔT_{f }= i.K_{f}m

= 1.0655 × 1.86 K kg mol^{-1} × 0.3264 mol kg^{-1}

= 0.65 K

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