To 500 cm^3 of water, 3.0 ×10^-3 kg of acetic acid is added. If 23 % of acetic acid is dissoci

Name: Problem 203, Chapter 9: Physical Chemistry for IIT JEE
Uploaded: 2022-01-31T15:57:18-08:00
Duration: 1 min 1 s
Channel: Narayan Hari
Description: To 500 cm^3 of water, 3.0 ×10^-3 kg of acetic acid is added. If 23 % of acetic acid is dissociated, what will be the depression in freezing point? Kf and density of water at 1.86 K kg-1 and 0.0997 g cm^-3 respectively. a. 0.186 K b. 0.228 K c. 0.371 K d. 0.555 K [IIT 2000]

step 1 In this problem given that alpha is 0 .23, i is equal to 1 plus alpha which is equal to 1 plus 0

Question

To $500 \mathrm{~cm}^{3}$ of water, $3.0 \times 10^{-3} \mathrm{~kg}$ of acetic acid is added. If $23 \%$ of acetic acid is dissociated, what will be the depression in freezing point? $\mathrm{K}_{\mathrm{f}}$ and density of water at $1.86 \mathrm{~K} \mathrm{~kg}-1$ and $0.0997 \mathrm{~g} \mathrm{~cm}^{-3}$ respectively. a. $0.186 \mathrm{~K}$ b. $0.228 \mathrm{~K}$ c. $0.371 \mathrm{~K}$ d. $0.555 \mathrm{~K}$ [IIT 2000]

   To $500 \mathrm{~cm}^{3}$ of water, $3.0 \times 10^{-3} \mathrm{~kg}$ of acetic acid is added. If $23 \%$ of acetic acid is dissociated, what will be the depression in freezing point? $\mathrm{K}_{\mathrm{f}}$ and density of water at $1.86 \mathrm{~K} \mathrm{~kg}-1$ and $0.0997 \mathrm{~g} \mathrm{~cm}^{-3}$
respectively.
a. $0.186 \mathrm{~K}$
b. $0.228 \mathrm{~K}$
c. $0.371 \mathrm{~K}$
d. $0.555 \mathrm{~K}$
[IIT 2000]

Physical Chemistry for IIT JEE

Atul Singha 1st Edition

Chapter 9, Problem 203

Instant Answer

Solved by Expert Narayan Hari

01/31/2022

Step 1

23, we can calculate the van't Hoff factor (i) using the formula: \[i = 1 + α = 1 + 0.23 = 1.23\] Show more…

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To $500 \mathrm{~cm}^{3}$ of water, $3.0 \times 10^{-3} \mathrm{~kg}$ of acetic acid is added. If $23 \%$ of acetic acid is dissociated, what will be the depression in freezing point? $\mathrm{K}_{\mathrm{f}}$ and density of water at $1.86 \mathrm{~K} \mathrm{~kg}-1$ and $0.0997 \mathrm{~g} \mathrm{~cm}^{-3}$ respectively. a. $0.186 \mathrm{~K}$ b. $0.228 \mathrm{~K}$ c. $0.371 \mathrm{~K}$ d. $0.555 \mathrm{~K}$ [IIT 2000]

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Key Concepts

Freezing Point Depression

This colligative property describes how the freezing point of a solvent decreases when a non?volatile solute is dissolved in it. The depression in the freezing point is directly proportional to the molality of the solution and the cryoscopic constant of the solvent, as given by the relation ?Tf = i·Kf·m.

Molality

Molality is the concentration unit expressed as the number of moles of solute per kilogram of solvent. It is particularly useful for colligative property calculations because it is independent of temperature, allowing for reliable comparisons across varying conditions.

Van’t Hoff Factor

The van’t Hoff factor represents the number of particles into which a solute dissociates in solution. It adjusts the contribution of each solute unit in colligative property calculations by accounting for partial or complete dissociation, which is critical for accurately determining properties like freezing point depression.

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