1. Tardigrade
  2. Question
  3. Chemistry
  4. Freezing point of 0.2 M KCN solution is -0.7° C. On adding 0.1 mole of Hg ( CN )2 to one litre of the 0.2 M KCN solution, the freezing point of the solution becomes -0.53° C due to the reaction Hg ( CN )2+ mCN - arrow[ Hg ( CN )] m +2 m - . What is the value of m ? (Assuming molality = molarity)

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Q. Freezing point of $0.2 \,M \,KCN$ solution is $-0.7^{\circ} C$. On adding $0.1$ mole of $Hg ( CN )_{2}$ to one litre of the $0.2$ $M\, KCN$ solution, the freezing point of the solution becomes $-0.53^{\circ} C$ due to the reaction $Hg ( CN )_{2}+ mCN ^{-}$ $\rightarrow[ Hg ( CN )]_{ m +2}^{ m -} .$ What is the value of $m ?$ (Assuming molality = molarity)

Solutions

Solution:

$0.7=2 \times K _{ f } \times 0.2$
$K _{ f }=\frac{0.7}{0.4}=\frac{7}{4}$
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Now, molarity of $K ^{+}=0.2 \, M$
Molarity of $CN ^{-}=(0.2-0.1 \,m ) M$
Molarity of complex $=0.1 \, M$
$0.53 = K _{ f }(0.2+0.2-0.1 \, m +0.1) $
$= K _{ f }(0.5-0.1 m )$
$0.53 =\frac{7}{4}(0.5-0.1 \, m )$
$2.1 =3.5-0.7 \, m $
$0.7 \,m =1.4 $
$m =2$