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Define boiling point and find out expression for the molecular mass of non-volatile solute from the elevation in boiling point. |
Answer» Solution :Elevation in boiling point. The boiling point of a liquid is that temperature at which its vapour PRESSURE becomes equal to the atmospheric pressure. The vapour pressure of a solution is lower than that of a pure solvent. This means that the solution has to be heated to a higher temperature, so that its vapour pressure becomes equal to atmospheric pressure. Therefore, the boiling point of a solvent rises on the addition of a nonvolatile solute and this is called elevation in boiling point.  The elevation in boiling point on the addition of a non-volatile solute to a solvent is shown in Fig. The curve AB represents vapour pressure of the solvent and curve CD represents the vapour pressure of the solution. It can be seen from the FIGURE that at point B on the curve AB, the vapour pressure of the solvent corresponds to atmospheric pressure, hence it corresponds to the boiling point `(T_(b)^(0))` of the pure solvent. At point D, the vapour of the solution becomes equal to atmospheric pressure and hence it corresponds to the boiling point (`T_b`) of thesolution. The difference `T_b - T_(b)^(2) = DELTA T_b` is called elevation in boiling point, because `T_b` is greater than `Delta T_(b)^(@)`. Elevation in boiling point, `Delta T_b = T_b - T_b^(@)`. It has been found out experimentally that the elevation in the boiling point `(Delta T_b)` of a solution is PROPORTIONAL to the molal concentration of the solution i.e. `Delta T_b propm orDelta T_b =K_b m,` where K is called molal elevation constant. This is defined as: ` Delta T_b =K_b xx m` `Delta T_b = (K_b xx W_b xx 1000 )/( W_A xx Delta T_b )` `or M_B = (K_b xx w_B xx 1000 )/( w_A xx Delta T_b)` `therefore` knowingalll otherfactors, `M_B` canbecalculated. |
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