Chemistry - Liquid Solution Question with Solution | TestHub

Question Explanation:

Find the freezing point of a 0.001N NH₄OH solution.

Concept:

This question is based on equivalent conductance, limiting molar conductance, and depression in freezing point.

Solution:

Limiting molar conductance:

${\Lambda }_{aq.N{H}_{4}Cl}^{\circ }=1.50\times 10^{-2}oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

${\Lambda }_{aq.NaCl}^{\circ }=1.26\times 10^{-2}oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

${\Lambda }_{aq.NaOH}^{\circ }=2.26\times 10^{-2}oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

NH₄Cl →NH₄⁺ + Cl^- ....(1)

NaCl → Na⁺ + Cl^-.....(2)

NaOH →Na⁺ + OH^- .....(3)

NH₄OH → NH₄⁺ + OH^-

${\Lambda }_m^{\infty }\left(N{H}_{4}OH\right)={\Lambda }_{NH\_4Cl}^{\circ }+{\Lambda }_{NaOH}^{\circ }-{\Lambda }_{NaCl}^{\circ }$

$=(1.50\times 10^{-2})+(2.26\times 10^{-2})-(1.26\times 10^{-2})$

$=2.50\times 10^{-2}oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

${\Lambda }_{eq}=\frac{{\Lambda }_m}{n-factor}$ n-factor = 1 for NH₄OH

On the basis of this ${\Lambda }_m=3.0\times 10^{-3}oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

Degree of dissociation $\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\infty }}=\frac{3.0\times 10^{-3}}{2.5\times 10^{-2}}=0.12$

Calculating van't Hoff factor: $i=1+(n-1)\alpha$. For NH₄OH, $n=2$ (dissociates into NH₄⁺ and OH⁻).

$i=1+(2-1)\alpha =1+\alpha$

$i=1+0.12=1.12$

Depression in freezing point $\Delta T_f=K_f\cdot m\cdot i$.

Here, $K_f=1.86Kkgmo{l}^{-1}$.

It is given that molality ($m$) = molarity ($M$).

Molarity (n-factor = 1) = Normality = 0.001 M.

So, $m=0.001molk{g}^{-1}$.

$\Delta T_f=(1.86\times 0.001)\times 1.12$

$\Delta T_f=0.0020832°C$

The freezing point of the solution is $T_f=0°C-\Delta T_f$.

$T_f=0°C-0.0020832°C\approx -0.002°C$

Final Answer: The correct option is (C).