(a) (i) Molarity of a solution is defined as the number of moles of solute present in one litre of the solution. (ii) Molal elevation constant $(K_b)$ is defined as the elevation of the boiling point of a solution when one mole of a non-volatile solute is dissolved in one kilogram of a volatile solvent. (b) Given; Mass of urea, $W_B=15\,\mathrm{g}$ Molar mass of urea, $M_{\text{urea}}=60\,\mathrm{g}$ The solution of urea in water is isotonic to that of glucose solution. $\pi_{\text{urea}} = \pi_{\text{glucose}}$ $C_{\text{urea}} RT = C_{\text{glucose}} RT$ $\frac{n_{\text{urea}} RT}{V} = \frac{n_{\text{glucose}}}{V} RT$ $\frac{15}{60} = \frac{W_{\text{glucose}}}{180}$ $W_{\text{glucose}} = \frac{15 \times 180}{60}$ $W_{\text{glucose}} = 45\,\mathrm{g}$ OR (a) According to Raoult's law, the partial pressure of a component is the product of vapour pressure of pure solvent and mole fraction of that component. When a solution shows deviation from Raoult's law over the complete range of concentration, the solution is known as a non-ideal solution. The vapour pressure of the non-ideal solution can be higher or lower than the vapour pressure predicted by Raoult's law. If the vapour pressure of the non-ideal solution is higher than the vapour pressure predicted by Raoult's law, the deviation is known as positive deviation. If the vapour pressure of the non-ideal solution is lower than the vapour pressure predicted by Raoult's law, the deviation is known as negative deviation. The reason for the deviation is molecular interactions, $A-B < A-A, B-B$ for positive deviation Here, A is the solute and B is the solvent so, A-B shows the interaction between solute and solvent. (b) The number of moles of glucose$= \frac{10}{180} = 0.056\,\mathrm{mol}$$\text{ Molality of solution }$ $= \frac{0.056\,\mathrm{mol}}{0.09\,\mathrm{kg}} = 0.62\,\mathrm{m}$If the density of the solution is $1.2\,\mathrm{g\,mL}^{-1}$, then the volume of the $100\,\mathrm{g}$ solution can be given as,$\frac{100\,\mathrm{g}}{1.2\,\mathrm{mL}^{-1}} = 83.3\,\mathrm{mL}$$= 83.33 \times 10^{-3}\,\mathrm{L}$$\text{ Molarity of solution }$ $= \frac{0.056\,\mathrm{mol}}{83.33 \times 10^{-3}\,\mathrm{L}} = 0.67\,\mathrm{M}$