Chapter 16

On dissolving \(0.5 \mathrm{~g}\) of a non-volatile non-ionic solute to \(39 \mathrm{~g}\) of benzene, its vapour pressure decreases from \(650 \mathrm{~mm} \mathrm{Hg}\) to \(640 \mathrm{~mm}\) \(\mathrm{Hg}\). The depression of freezing point of Benzene (in \(\mathrm{K}\) ) upon addition of the solute is (Given data : Molar mass and the molal freezing point depression constant of benzene are \(78 \quad \mathrm{~g} \quad \mathrm{~mol}^{-1} \quad\) and \(5.12 \quad \mathrm{~K} \quad \mathrm{~kg} \quad \mathrm{~mol}^{-1}\), respectively)

75.2 \(\mathrm{g}\) of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) (phenol) is dissolved in a solvent of \(K_{f}=14\). If the depression in freezing point is \(7 \mathrm{~K}\) then find the \(\%\) of phenol that dimerises.

A bottle of commercial sulphuric acid (density \(1.787 \mathrm{~g} / \mathrm{mL}\) ) is labelled as 86 percent by weight. What is the molarity of the acid? What volume of the acid has to be used to make 1 litre of \(0.2 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\)

To \(500 \mathrm{~cm}^{3}\) of water, \(3.0 \times 10^{-3} \mathrm{~kg}\) of acetic acid is added. If \(23 \%\) of acetic acid is dissociated, what will be the depression in freezing point? \(K_{f}\) and density of water are \(1.86 \mathrm{~K} \mathrm{~kg}^{1} \mathrm{~mol}^{1}\) and \(0.997 \mathrm{~g} \mathrm{~cm}^{3}\), respectively.

What is the molarity and molality of a \(13 \%\) solution (by weight) of sulphuric acid with a density of \(1.02 \mathrm{~g} / \mathrm{mL} ?\) To what volume should \(100 \mathrm{~mL}\) of this acid be diluted in order to prepare a \(1.5 \mathrm{~N}\) solution?

A solution of a non-volatile solute in water freezes at \(-0.30^{\circ} \mathrm{C}\). The vapour pressure of pure water at \(298 \mathrm{~K}\) is \(23.51 \mathrm{~mm} \mathrm{Hg}\) and \(K_{f}\) for water is \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} .\) Calculate the vapour pressure of this solution at \(298 \mathrm{~K}\).

Addition of \(0.643 \mathrm{~g}\) of a compound to \(50 \mathrm{~mL}\). of benzene (density : \(0.879 \mathrm{~g} / \mathrm{mL}\).) lowers the freezing point from \(5.51^{\circ} \mathrm{C}\) to \(5.03^{\circ} \mathrm{C}\). If \(K_{f}\) for benzene is \(5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\), calculate the molecular weight of the compound.

The degree of dissociation of calcium nitrate in a dilute aqueous solution, containing \(7.0 \mathrm{~g}\). of the salt per \(100 \mathrm{~g}\) of water at \(100^{\circ} \mathrm{C}\) is \(70 \%\). If the vapour pressure of water at \(100^{\circ} \mathrm{C}\) is \(760 \mathrm{~mm}\), calculate the vapour pressure of the solution.

The vapour pressure of pure benzene at a certain temperature is 640 \(\mathrm{mm}\) Hg. A non-volatile non-electrolyte solid weighing \(2.175 \mathrm{~g}\) is added to \(39.0 \mathrm{~g}\) of benzene. The vapour pressure of the solution is 600 \(\mathrm{mm} \mathrm{Hg}\). What is the molecular weight of the solid substance?

Given that \(\Delta T_{f}\) is the depression in freezing point of the solvent in a solution of a non-volatile solute of molality, \(m\), the quantity \(\operatorname{Lim}_{m \rightarrow 0}\left(\Delta T_{f} / m\right)\) is equal to \(\ldots \ldots \ldots . .\)

In the depression of freezing point experiment, it is found that the (a) vapour pressure of the solution is less than that of pure solvent (b) vapour pressure of the solution is more than that of pure solvent (c) only solute molecules solidify at the freezing point (d) only solvent molecules solidify at the freezing

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution \(\mathrm{M}\) is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is \(0.9\) Given : Freezing point depression constant of water \(\left(K_{f}^{\text {water }}\right)\) $$ =1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Freezing point depression constant of ethanol ( \(\left.K_{f}{\underline{\phantom{xx}}}^{\text {ethanol }}\right)\) $$ =2.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Boiling point elevation constant of water \(\left(K_{b}^{\text {water }}\right)\) \(=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Boiling point elevation constant of ethanol \(\left(K_{b}^{\text {ethanol }}\right)=1.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Standard freezing point of water \(=273 \mathrm{~K}\) Standard freezing point of ethanol \(=155.7 \mathrm{~K}\) Standard boiling point of water \(=373 \mathrm{~K}\) Standard boiling point of ethanol \(=351.5 \mathrm{~K}\) Vapour pressure of pure water \(=32.8 \mathrm{~mm} \mathrm{Hg}\) Vapour pressure of pure ethanol \(=40 \mathrm{~mm} \mathrm{Hg}\) Molecular weight of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\) Molecular weight of ethanol \(=46 \mathrm{~g} \mathrm{~mol}^{-1}\) In answering the following questions, consider the solution to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The freezing point of the solution \(\mathrm{M}\) is (a) \(268.7 \mathrm{~K}\) (b) \(268.5 \mathrm{~K}\) (c) \(234.2 \mathrm{~K}\) (d) \(150.9 \mathrm{~K}\)

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution \(\mathrm{M}\) is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is \(0.9\) Given : Freezing point depression constant of water \(\left(K_{f}^{\text {water }}\right)\) $$ =1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Freezing point depression constant of ethanol ( \(\left.K_{f}{\underline{\phantom{xx}}}^{\text {ethanol }}\right)\) $$ =2.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Boiling point elevation constant of water \(\left(K_{b}^{\text {water }}\right)\) \(=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Boiling point elevation constant of ethanol \(\left(K_{b}^{\text {ethanol }}\right)=1.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Standard freezing point of water \(=273 \mathrm{~K}\) Standard freezing point of ethanol \(=155.7 \mathrm{~K}\) Standard boiling point of water \(=373 \mathrm{~K}\) Standard boiling point of ethanol \(=351.5 \mathrm{~K}\) Vapour pressure of pure water \(=32.8 \mathrm{~mm} \mathrm{Hg}\) Vapour pressure of pure ethanol \(=40 \mathrm{~mm} \mathrm{Hg}\) Molecular weight of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\) Molecular weight of ethanol \(=46 \mathrm{~g} \mathrm{~mol}^{-1}\) In answering the following questions, consider the solution to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The vapour pressure of the solution \(\mathrm{M}\) is (a) \(39.3 \mathrm{~mm} \mathrm{Hg}\) (b) \(36.0 \mathrm{~mm} \mathrm{Hg}\) (c) \(29.5 \mathrm{~mm} \mathrm{Hg}\) (d) \(28.8 \mathrm{~mm} \mathrm{Hg}\)

\(1.22 \mathrm{~g}\) of benzoic acid is dissolved in \(100 \mathrm{~g}\) of acetone and \(100 \mathrm{~g}\) of benzene separately. Boiling point of the solution in acetone increases by \(0.17^{\circ} \mathrm{C}\), while that in the benzene increases by \(0.13^{\circ} \mathrm{C}\); \(K_{b}\) for acetone and benzene is \(1.7 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) and \(2.6 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) respectively. Find molecular weight of benzoic acid in two cases and justify your answer.

Match the following if the molecular weights of \(X, Y\) and \(Z\) are same. $$ \begin{array}{cll} \hline & \text { Boiling Point } & \boldsymbol{K}_{b} \\ \hline X & 100 & 0.63 \\ \hline Y & 27 & 0.53 \\ \hline Z & 253 & 0.98 \end{array} $$

Nitrobenzene is formed as the major product along with a minor product in the reaction of benzene with a hotmixture of nitric acid and sulphuric acid. The minorproduct consists of carbon : \(42.86 \%\), hydrogen : \(2.40 \%\), nitrogen : \(16.67 \%\), and oxygen: \(38.07 \%\) (i) Calculate the empirical formula of the minor product. (ii) When \(5.5 \mathrm{~g}\) of the minor product is dissolved in \(45 \mathrm{~g}\) of benzene, the boiling point of the solution is \(1.84^{\circ} \mathrm{C}\) higher than that of pure benzene. Calculate the molar mass of the minor product and determine its molecular and structural formula. (Molal boiling point elevation constant of benzene is \(\left.2.53 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} .\right)\)