Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15°C. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at −0.

Question

Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15°C. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at −0.8°C. The solubility product of  formed is  at 298 K. (Nearest integer). Given:  and . Assume molality = molarity.

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Strategy:
 First, determine the initial concentration of Lead nitrate. Then, when  is added,  precipitates. Use the final freezing point to find the remaining concentration of ions and solve for the solubility product ().

Step 1: Initial Concentration of Lead nitrate
 (initial) , .
 dissociates into 3 ions ().

Initial: , .

Step 2: Addition of } \ce{NaCl} 	ext{
Added 0.2 mol  to 1 L .
Due to the stoichiometry of  (),  reacts with , meaning most of it precipitates. Let  be the concentration of

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