Question

Calculate the freezing point of a solution containing 0.5 g kcl (molar mass = 74.5 g/mol) dissolved in 100 g water, assuming kcl to be 92% ionized.

Answer 1

Explanation:

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Answer 2

Freezing point of the solution after the addition of KCl = -0.23°K

Explanation:

Depression in freezing point

$\Delta T _f =1000 \times k_f \times \frac{w_2}{M_2}$

• Form the given values,

• Degree of dissociation $\alpha =0.92$  (The percentage of dissociation is 92%)

• After dissociation, Total no. of moles$=\alpha+1=i$

• $i= \frac{calculated \,molar \,mass}{theoretical \,molar \,mass}$

• $\Rightarrow 1.92=\frac{74.5}{theoretical \,molar \,mass}$

• $\Rightarrow theoretical\,molar\,mass=38.802$

• $\therefore Depression\, in \,freezing \,point=1000\times 1.86 \times \frac{0.5}{38.802} \times 100=0.23K$

• We know,

• Depression in freezing point $\Delta T_f=$ Initial freezing point - freezing point after addition of solute

• $\Rightarrow$ 0.23=0-freezing point after addition of solute

• $\therefore$          freezing point after addition of solute = -0.23K