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JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
Solid Lead nitrate is dissolved in \(1\,litre\) of water. The solution was found to boil at \(100.15^{\circ}\,C\). When \(0.2\,mol\) of \(NaCl\) is added to the resulting solution, it was observed that the solution froze at \(-0.8^{\circ}\,C\). The solutbility product of \(PbCl _2\) formed is \(...........\times 10^{-6}\) at \(298\,K\). (Nearest integer) Given : \(K _{ b }=0.5\,K\,kg\,mol ^{-1}\) and \(K _{ f }=1.8\,kg\,mol ^{-1}\). Assume molality to be equal to molarity in all cases.
- A \(13\)
- B \(12\)
- C \(11\)
- D \(10\)
Answer & Solution
Correct Answer
(A) \(13\)
Step-by-step Solution
Detailed explanation
Let a mole \(Pb \left( NO _3\right)_2\) be added \(Pb \left( NO _3\right)_2 \rightarrow Pb ^{2+}+2 NO _3^{-}\) a \(\quad\quad\quad\quad\quad\quad\) a \(\quad\qquad 2 a\) \(\Delta T _{ b }=0.15=0.5[3 a ] \Rightarrow a =0.1\)…
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