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JEE Main 202310 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
10 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2023 · 10 April · Morning Shift · Chemistry · Solutions

If the degree of dissociation of aqueous solution of weak monobasic...

If the degree of dissociation of aqueous solution of weak monobasic acid is determined to be 0.30 . 3, then the observed freezing point will be _____ % higher than the expected/theoretical freezing point. (Nearest integer).

Correct answer

0

Step-by-step solution

HA1−α⇋H+α+A−α\mathop{HA}_{1 - \alpha} ⇋ \mathop{{H}^{+}}_{\alpha} + \mathop{{A}^{-}}_{\alpha}

Van't Hoff factor, i=1+α(n−1)i = 1 + \alpha\left(n - 1\right)

=1+0.3(2−1)= 1 + 0 . 3\left(2 - 1\right)

=1+0.3= 1 + 0 . 3

=1.3= 1 . 3

∆Tf∆ {T}_{f} without considering dissociation =kf×= {k}_{f} \times molality

∆Tf∆ {T}_{f} without considering dissociation =ikf×= {ik}_{f} \times molality

=ΔTf′= {\Delta T}_{f}^{'}

ΔTf′−ΔTf=0.3{\Delta T}_{f}^{'} - {\Delta T}_{f} = 0 . 3 (kf×{k}_{f} \times molality)

ΔTf1−ΔTfΔTf=0.3(kf×molality)kf×molality=0.3\dfrac{{\Delta T}_{f}^{1} - {\Delta T}_{f}}{{\Delta T}_{f}} = \dfrac{0 . 3\left({k}_{f} \times molality\right)}{{k}_{f} \times molality} = 0 . 3

% value =30= 30

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