Answered

Compared to the freezing point of 1.0 M KCl(aq) at standard pressure, the freezing point of 1.0 M CaCl2(aq) at standard pressure is

(1) lower

(2) higher

(3) the same

Verified Answer :

Answer : The correct option is, (1) lower

Explanation :

Formula used for lowering in freezing point :

\(\Delta T_f=i\times k_f\times m\)

where,

\(\DeltaT_b\) = change in freezing point

\(k_b\) = freezing point constant

m = molality

i = Van't Hoff factor

According to the question, we conclude that the molality of the given solutions are the same. So, the freezing point depends only on the Van't Hoff factor.

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) The dissociation of \(KCl\) will be,

\(KCl\rightarrow K^++Cl^-\)

So, Van't Hoff factor = Number of solute particles = 1 + 1 = 2

(b) The dissociation of \(CaCl_2\) will be,

\(CaCl_2\rightarrow Ca^{2+}+2Cl^-\)

So, Van't Hoff factor = Number of solute particles = 1 + 2 = 3

The freezing point depends only on the Van't Hoff factor. That means higher the Van't Hoff factor, lower will be the freezing point and vice-versa.

Thus, compared to the freezing point of 1.0 M \(KCl(aq)\) at standard pressure, the freezing point of 1.0 M \(CaCl_2(aq)\) at standard pressure is lower.

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