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[${K_F} = {2.0^0}mo{l^{ - 1}} kg$, molarity = molality]

a.) ${3.2^0}$

b.) ${1.6^0}$

c.) ${0.8^0}$

d.) ${5.0^0}$

Answer
Verified

It can be calculated by the formula as-

$\Delta {T_f}$ = molality × ${K_f}$ × i

Where $\Delta {T_f}$ is the depression in freezing point

${K_f}$ is the molal depression constant

And I is the van't hoff factor

In the question, we have been given that 25 ml of KCl react with 20 ml of 1 M $AgN{O_3}$ solution.

Thus, volume of KCl = 25 ml

Molarity of KCl = M

Volume of $AgN{O_3}$ solution = 1 M

Molarity of $AgN{O_3}$ solution = 1 M

When the reaction occurred; we know that during neutralization

${M_1}{V_1} = {M_2}{V_2}$

Where ${M_1}$ is the molarity of KCl solution

${M_2}$ is the molarity of $AgN{O_3}$ solution

${V_1}$ is the volume of KCl solution

And ${V_2}$ is the volume of $AgN{O_3}$ solution

Thus, by putting the values, we can find molarity

${M_1}$× 25 = 1 × 20

Therefore, ${M_1} = \dfrac{{1×20}}{{25}}$

${M_1}$ = 0.8 M

Further, we are given that molarity is equal to molality.

Thus, molality = molarity = 0.8 molal

We have the formula for finding the depression in freezing point as-

$\Delta {T_f}$ = molality × ${K_f}$ × i

We will first find value if I which is Von’t hoff factor.

i = (1+α)

As the dissociation is 100%. So, the value of α will be 1.

Thus, i = (1+1)

i = 2

So, now we have all the variables and we can find out the value of depression in freezing point as-

$\Delta {T_f}$ = molality × ${K_f}$ × i

$\Delta {T_f}$ = 0.8 M × 2.0 × 2

$\Delta {T_f}$ = ${3.2^0}$

The depression in freezing point is a colligative property and depends upon the number of particles present.

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