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)

Question

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) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

0.7=2 × K f × 0.2 K f =(0.7/0.4)=(7/4) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/2a9249ac1ec2d63fade3057b972f8eb9-.png" /> 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 =(7/4)(0.5-0.1 m ) 2.1 =3.5-0.7 m 0.7 m =1.4 m =2

Q. Freezing point of 0.2MKCN solution is −0.7∘C. On adding 0.1 mole of Hg(CN)2​ to one litre of the 0.2 MKCN solution, the freezing point of the solution becomes −0.53∘C due to the reaction Hg(CN)2​+mCN− →[Hg(CN)]m+2m−​. What is the value of m? (Assuming molality = molarity)

0.7=2×Kf​×0.2 Kf​=0.40.7​=47​ Now, molarity of K+=0.2M Molarity of CN−=(0.2−0.1m)M Molarity of complex =0.1M 0.53=Kf​(0.2+0.2−0.1m+0.1) =Kf​(0.5−0.1m) 0.53=47​(0.5−0.1m) 2.1=3.5−0.7m 0.7m=1.4 m=2