The freezing point of an aqueous solution of KCN containing 0.2 mole / kg water was − 0.8 0 ∘ C . On adding 0.1 mole of Hg ( CN ) 2 in the solution containing 1 kg of water, the freezing point of th

Question

The freezing point of an aqueous solution of KCN containing 0.2 mole / kg water was − 0.8 0 ∘ C . On adding 0.1 mole of Hg ( CN ) 2 ​ in the solution containing 1 kg of water, the freezing point of the solution was − 0. 6 ∘ C . Assuming that the complex is formed according to the following equation Hg ( CN ) 2 ​ + m C N − → Hg ( CN ) m + 2 m − ​ and Hg ( CN ) 2 ​ is the limiting reactant, the value of m is

TestHub · Accepted Answer

Δ T f ​ = K f ​ × m For KCN solution: 0.80 = K f ​ × 0.2 × 2.. … ( 1 ) Hg(CN)₂ + mCN⁻ ⇌ Hg(CN)₂₊ₘᵐ⁻ 0.1 mole 0.2 mole 0 0 (0.2 - 0.1m) mole 0.1 mole Final effective molality = 0.2 − 0.1 m + 0.1 + 0.2 = 0.5 − 0.1 m Now, 0.60 = K f ​ × ( 0.5 − 0.1 m ) . …… ( 2 ) From (1) & (2) we get, m = 2

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JEE Main - 1 year (12th class)