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Standard XII

Chemistry

Van't Hoff Factor

0.6 ml of gla...

Question

$0.6 m l$ of glacial acetic acid with density $1.06 g / m l$ is dissolved in $1 k g$ water and the solution froze at $- {0.0205}^{\circ} C$ . Calculate van't Hoff factor:

$K_{f}$ for water is $1.86 K k g m o l^{- 1}$ .

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Solution

Mass of acetic acid $= 0.6 m l \times 1.06 g m l^{- 1} = 0.636$

No. of moles of acetic acid $= \frac{0636}{60} = 0.0106 m o l$

Using the equation :-

$Δ T_{f} = 1.86 \times 0.0106 = 0.0197 K$

van't Hoff Factor $= \frac{o b s e r v e d f r o z i n g p o i n t}{c a l c u l a t e f r o z i n g p o i n t}$

$= \frac{0.0205}{0.0197} = 1.04$

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Similar questions

0.6

ml of glacial acetic acid with density

1.06

L^{- 1}

is dissolved in

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- {0.0205}^{o}

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QUESTION 2.33

19.5 g of $C H_{2} F C O O H$ is dissolved in 500g of water. The depression in the freezing point of water observed in 1.0 C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.
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, at. wt. Na = 23, Cl = 35.5)

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0.6 ml of glacial acetic acid with density 1.06 g/ml is dissolved in 1 kg water and the solution froze at −0.0205∘C. Calculate van't Hoff factor: Kf for water is 1.86 K kg mol−1.

Mass of acetic acid =0.6ml×1.06gml−1=0.636No. of moles of acetic acid =063660=0.0106molUsing the equation :-ΔTf=1.86×0.0106=0.0197Kvan't Hoff Factor =observedfrozingpointcalculatefrozingpoint=0.02050.0197=1.04

$0.6 m l$ of glacial acetic acid with density $1.06 g / m l$ is dissolved in $1 k g$ water and the solution froze at $- {0.0205}^{\circ} C$ . Calculate van't Hoff factor:

$K_{f}$ for water is $1.86 K k g m o l^{- 1}$ .