Example 1.8

Example . D T b = K b × m = . K kg mol – × . mol kg – = . K Since water boils at . K at . bar pressure, therefore, the boiling point of solution will be . + . = . K. The boiling point of benzene is . K. When . g of a non-volatile solute was dissolved in g of benzene, the boiling point is raised to . K. Calculate the molar mass of the solute. K b for benzene is . K kg mol – The elevation ( D T b ) in the boiling point = . K – . K = . K Substituting these values in expression ( . ) we get M = – – . K kg mol × . g × g kg . K × g = g mol – Therefore, molar mass of the solute, M = g mol – Fig. . : Diagram showing D T f , depression of the freezing point of a solvent in a solution. . . Depression of Freezing Point Depression Constant or Cryoscopic Constant . The unit of K f is K kg mol - . Values of K f for some common solvents are listed in Table . . If w gram of the solute having molar mass as M , present in w gram of solvent, produces the depression in freezing point D T f of the solvent then molality of the solute is given by the equation ( . ). m = w / / M ( . ) Substituting this value of molality in equation ( . ) we get: D T f = f / /  K M D T f = f × K M ( . ) M = f f ×  K T ( . ) Thus for determining the molar mass of the solute we should know the quantities w , w , D T f , along with the molal freezing point depression constant. The values of K f and K b