(a) (i) Define saturated solution. (ii) Tha solubility of KNO\(_3\) at 20°C was 3.00 mol dm\(^{-3}\) If 67.0g of KNO\(_3\) was added to 250 cm\(^{-3}\) of w...

(a) Saturated solution

• (i) A saturated solution is one that contains the maximum amount of dissolved solute that it can hold at a given temperature, in the presence of undissolved solute (any further solute added will not dissolve).
• (ii) Solubility of KNO\(_3\) at 20 degrees Celsius \(= 3.00\ \text{mol dm}^{-3}\). Amount added: molar mass of KNO\(_3 = 101.0\), so \[n = \frac{67.0}{101.0} = 0.663\ \text{mol}\ \text{in}\ 250\ \text{cm}^3\ (0.250\ \text{dm}^3).\] Concentration formed \[= \frac{0.663}{0.250} = 2.65\ \text{mol dm}^{-3}.\] Since \(2.65\ \text{mol dm}^{-3} < 3.00\ \text{mol dm}^{-3}\) (the solubility), the solution is unsaturated, i.e. it is not saturated at 20 degrees Celsius.

(b) Bonding

• (i) In an ordinary covalent bond each of the two bonded atoms contributes one electron to the shared pair, whereas in a dative (co-ordinate) bond both electrons of the shared pair are supplied by only one of the atoms.
• (ii) Sugar and common salt do not conduct in the solid state because sugar is a covalent (molecular) solid with no ions or free electrons, and although common salt is ionic, in the solid its ions are held rigidly in the lattice and are not free to move; conduction needs mobile charge carriers.
• (iii) I. Hydrogen fluoride: hydrogen bonding (plus weaker dipole-dipole/van der Waals). II. Argon: weak van der Waals (dispersion) forces only.
• (iv) The hydrogen bond is stronger in O-H----N. Reason: oxygen is more electronegative than sulphur, so the O-H bond is more polar and the hydrogen carries a greater partial positive charge, giving a stronger attraction to the nitrogen lone pair.

(c) Gas laws

• (i) Dalton's law of partial pressures: in a mixture of gases that do not react, the total pressure is equal to the sum of the partial pressures of the individual gases, each partial pressure being the pressure the gas would exert if it alone occupied the whole volume at the same temperature.
• (ii) Pressure of dry CO\(_2\) \(= 700 - 15 = 685\ \text{mmHg}\); \(V_1 = 200\ \text{cm}^3\), \(T_1 = 18 + 273 = 291\ \text{K}\). At s.t.p. \(P_2 = 760\ \text{mmHg}\), \(T_2 = 273\ \text{K}\). Using \(\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\): \[V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{685 \times 200 \times 273}{291 \times 760} = 169.1\ \text{cm}^3.\] The volume of the dry gas at s.t.p. is about 169 cm\(^3\).