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Question:

The plot given shows P−T curves (where P is the pressure and T is the temperature) for two solvents X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents. On addition of an equal number of moles a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of the boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is:

Solution:

From graph,For solvent 'X'; ΔTb(X)=2ΔTb(X)=i×mNaCl×Kb(X).. (1)For solvent 'Y'; ΔTb(Y)=1ΔTb(Y)=i×mNaCl×Kb(Y)(2)Equation(1)/(2)⇒Kb(X)Kb(Y)=2For solute S2(S)⇌S21−αα/2Now,i=(1−α/2)ΔTb(X)(S)=(1−α12)Kb(X)mΔTb(Y)(S)=(1−α22)Kb(Y)mGivenΔTb(X)(S)=3ΔTb(Y)(S)(1−α12)Kb(X)=3×(1−α22)×Kb(Y)2(1−α12)=3(1−α22)–––−3α2=0.7So, on substituting the value of α2 in 3 equation we get,α1=0.05.