TS EAMCET · Chemistry · Solutions

The freezing point depression of $0.001 M$ of $A_{x} B_{y} [Fe {(CN)}_{6}]$ is $5.58 \times 10^{- 3} K$ . If the oxidation state of $Fe$ is $+ 2$ and $K_{f} = 1.86 K {kgmol}^{- 1}$ , then the total number of possibilities for different types of $A$ and $B$ cations are

A $1$
B $2$
C $3$
D $4$

Verified Solution

Answer & Solution

Correct Answer

(B) $2$

Step-by-step Solution

Detailed explanation

According to depression in freezing point:-∆Tf=iKfMAxByFeCN6 →XA+ + YB+ + FeCN62-i=∆TfKf×M∆Tf = 5.58×10-3Kf = 1.86 and M= 0.001=1×10-3So, i= 5.58&…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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The freezing point depression of 0.001 M of AxByFeCN6 is 5.58×10-3 K. If the oxidation state of Fe is +2 and Kf=1.86 K kgmol-1, then the total number of possibilities for different types of A and B cations are

Step-by-step Solution

The freezing point depression of $0.001 M$ of $A_{x} B_{y} [Fe {(CN)}_{6}]$ is $5.58 \times 10^{- 3} K$ . If the oxidation state of $Fe$ is $+ 2$ and $K_{f} = 1.86 K {kgmol}^{- 1}$ , then the total number of possibilities for different types of $A$ and $B$ cations are