Question

To $500c{m^3}$ of water, $3.0 \times {10^{ - 3}}kg$ of acetic acid is added. If $23\%$of acetic acid is dissociated, what will be the depression in freezing point? ${k_f}$ and the density of water are $1.86Kk{g^{ - 1}}$ and $0.997gc{m^{ - 3}}$,respectively.
a. $0.186K$
b. $0.228K$
c. $0.372K$
d. $0.556K$

Answer 1

To answer this question we will use 2 formulas. First we will use the formula for depression in freezing point .After substituting the values in that formula we will determine the number of particles after dissociation. And then using the van't-Hoff factor we will get to the correct answer.
FORMULA USED:
$\Delta {T_f} = \dfrac{{1000 \times w \times {k_f}}}{{m \times W}}$
Where $\Delta {T_f}$ is depression of freezing point
W is weight of acetic acid
W= weight of water
And ${k_f}$is a constant with value $1.86Kk{g^{ - 1}}$

Complete answer:
We know that Density of water is $0.997gc{m^{ - 3}}$
Let weight of water be W,
Hence,
$W = 500 \times 0.997g$
Or, $W = 498.5g$
Also let the weight of acetic acid be w,
Hence,
$w = 3.0 \times {10^{ - 3}}kg$
Or, $w = 3.0g$
We also know that the formula of elevation in freezing point is:
$\Delta {T_f} = \dfrac{{1000 \times w \times {k_f}}}{{m \times W}}$
Also the molar weight of acetic acid is 60m
Hence substituting the values in the above formula we get:
${\left( {\Delta {T_f}} \right)_{cal}} = \dfrac{{1000 \times 3 \times 1.86}}{{60 \times 498.5}}$
Or, ${\left( {\Delta {T_f}} \right)_{cal}} = 0.186$
Now we will consider the line in the question that states that $23\%$of acetic acid is dissociated,
Hence

At $t = 0$ $1$ $0$ $0$
At eqm $1 - \alpha$ $\alpha$ $\alpha$
Hence the total number of particles =$1 - \alpha + \alpha + \alpha = 1 + \alpha$
Also $\alpha$ for acetic acid will be $\dfrac{{23}}{{100}} = 0.23$
hence the total number of particles will be $= 1 + 0.23 = 1.23$
By using van’t-Hoff factor :
$\dfrac{{{{\left( {\Delta {T_f}} \right)}_{obs}}}}{{{{\left( {\Delta {T_f}} \right)}_{cal}}}} = \dfrac{{1.23}}{1}$
Or, ${\left( {\Delta {T_f}} \right)_{obs}} = 1.23 \times 0.186K$
Or, ${\left( {\Delta {T_f}} \right)_{obs}} = 0.228K$
Hence the depression in freezing point will be ${\left( {\Delta {T_f}} \right)_{obs}} = 0.228K$

Hence the correct answer to this question will be option b.

Note:
Yet another way of solving would be by using molality:
Number of moles of acetic acid$= \dfrac{{3g}}{{60gmo{l^{ - 1}}}} = 0.05$
Mass of water$= {m_1} = 500c{m^3} \times 0.997gc{m^{ - 3}}$
Or, ${m_1} = 0.4985kg$
Molality of acetic acid $m = \dfrac{{{n_2}}}{{{m_1}}} = \dfrac{{0.05mol}}{{0.4985kg}} = 0.1003molk{g^{ - 1}}$
Then similarly we will find the number of particles i.e,. $1 - \alpha + \alpha + \alpha = 1 + \alpha$. We will name this as i.
Then using the formula : $\Delta {T_f} = i{K_f}m$
We will get the answer as: $\Delta {T_f} = (1 + \alpha ){K_f}m = (1 + 0.23)(1.86\;K\;Kg\;mo{l^{ - 1}})(0.1003\;mol\;K{g^{ - 1}})$
Or, $\Delta {T_f} = 0.228K$