What is boiling point of solution containing 7.45 g of KCl in 500 g of water, assuming 80 % dissociation of KCl ? (molar mass of KCl = 74.5 g/mole , k_b for water = 0.52 K/molal )

Step-by-step solution

Molality of the solution is given by: m = w M W( in kg ) = 7.45 74.5 0.5 = 0.2 m For KCl , the dissociation reaction is KCl K ^+ + Cl ^- . The van't Hoff factor i is: i = 1 + (n - 1) Since n = 2 and = 0.8 : i = 1 + (2 - 1) 0.8 = 1.8 The elevation in boiling point is: T_b = i K_b m T_b = 1.8 0.52 0.2 = 0.1872 ^ C The boiling point of the solution is: T_b = 100 ^ C + T_b = 100 ^ C + 0.1872 ^ C = 100.1872 ^ C Answer: 100.1872 ^ C