Answer
Verified
394.8k+ views
Hint: Van’t Hoff factor is the measure of particles present after dissociation divided by the number of particles present initially. It is represented by ‘i’ and has the formula $\dfrac{\text{particles after dissociation/association}}{\text{particles before dissociation/association}}$ . If van’t Hoff factor is greater than 1, then dissociation occurs. If its value is less than 1, then association occurs and if van’t Hoff factor is equal to 1, then neither dissociation nor association occurs.
Complete answer:
Step (1)- Write the formula: Considering as an ideal solution, the depression in freezing point depends on the solute concentration that is e expressed as $\vartriangle {{\text{T}}_{\text{f}}}={{\text{K}}_{\text{f}}}\times \text{m}\times \text{i}$; where
- $\vartriangle {{\text{T}}_{\text{f}}}$ is the freezing-point depression expressed as${{\text{T}}_{\text{f}}}$ (pure solvent) − ${{\text{T}}_{\text{f}}}$(solution).
- ${{\text{K}}_{\text{f}}}$ is the cryoscopic constant which only depends on the properties of the solvent.
- $\text{m}$ is the molality which is represented as moles solute per kilogram of solvent or $\dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
- $\text{i}$ is van't Hoff factor.
Step (2)- Find molality by finding the moles of acetic acid; moles are defined as the weight of substance per molar mass of it. Its formula is$\text{moles=}\dfrac{\text{mass of substance}}{\text{molar mass}}$. Molar mass of acetic acid $\left( \text{C}{{\text{H}}_{3}}\text{COOH} \right)$:
- Atomic mass of $\text{C}$ is 12 grams.
- Atomic mass of $\text{H}$ is 1 gram.
- Atomic mass of $\text{O}$ is 16 grams.
Molar Mass is $\left[ \left( 12\times 2 \right)+\left( 1\times 4 \right)+\left( 16\times 2 \right) \right]$ or 60 grams.
Step (3)- Write the values given in the question; mass of acetic acid is given as 0.3 grams and mass of benzene (solvent) = 30g, $\vartriangle {{\text{T}}_{\text{f}}}=0.45\text{K}$ and ${{\text{K}}_{\text{f}}}$ for benzene = $5.12\text{ K kg mol}{{\text{e}}^{-1}}$.
Step (4)- The molality becomes $\dfrac{0.3\times \text{1000}}{60\times 30}$. The term is divided by 1000 because we have to find per ‘kg’ of solvent. The value of molality is 0.167 m.
Step (5)- Put the value of m in the formula $\vartriangle {{\text{T}}_{\text{f}}}={{\text{K}}_{\text{f}}}\times \text{m}\times \text{i}$; so, $0.45=5.12\times 0.167\times \text{i}$, van’t Hoff factor is 0.527.
The correct answer is van’t Hoff factor is 0.527.
Note: The value of Van’t Hoff factor obtained is nearing the actual experimental value of i for acetic acid. Van’t Hoff factor is nearly 0.5. [acetic acid has the tendency to dimerize (two molecules combine), so the number of particles after association is 1 and before association is 2. The van’t Hoff factor will be $\dfrac{1}{2}=0.5$]. So, the correct answer can be interpreted easily just by knowing that acetic acid will dimerize. The dimerization of acetic acid looks like
Complete answer:
Step (1)- Write the formula: Considering as an ideal solution, the depression in freezing point depends on the solute concentration that is e expressed as $\vartriangle {{\text{T}}_{\text{f}}}={{\text{K}}_{\text{f}}}\times \text{m}\times \text{i}$; where
- $\vartriangle {{\text{T}}_{\text{f}}}$ is the freezing-point depression expressed as${{\text{T}}_{\text{f}}}$ (pure solvent) − ${{\text{T}}_{\text{f}}}$(solution).
- ${{\text{K}}_{\text{f}}}$ is the cryoscopic constant which only depends on the properties of the solvent.
- $\text{m}$ is the molality which is represented as moles solute per kilogram of solvent or $\dfrac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
- $\text{i}$ is van't Hoff factor.
Step (2)- Find molality by finding the moles of acetic acid; moles are defined as the weight of substance per molar mass of it. Its formula is$\text{moles=}\dfrac{\text{mass of substance}}{\text{molar mass}}$. Molar mass of acetic acid $\left( \text{C}{{\text{H}}_{3}}\text{COOH} \right)$:
- Atomic mass of $\text{C}$ is 12 grams.
- Atomic mass of $\text{H}$ is 1 gram.
- Atomic mass of $\text{O}$ is 16 grams.
Molar Mass is $\left[ \left( 12\times 2 \right)+\left( 1\times 4 \right)+\left( 16\times 2 \right) \right]$ or 60 grams.
Step (3)- Write the values given in the question; mass of acetic acid is given as 0.3 grams and mass of benzene (solvent) = 30g, $\vartriangle {{\text{T}}_{\text{f}}}=0.45\text{K}$ and ${{\text{K}}_{\text{f}}}$ for benzene = $5.12\text{ K kg mol}{{\text{e}}^{-1}}$.
Step (4)- The molality becomes $\dfrac{0.3\times \text{1000}}{60\times 30}$. The term is divided by 1000 because we have to find per ‘kg’ of solvent. The value of molality is 0.167 m.
Step (5)- Put the value of m in the formula $\vartriangle {{\text{T}}_{\text{f}}}={{\text{K}}_{\text{f}}}\times \text{m}\times \text{i}$; so, $0.45=5.12\times 0.167\times \text{i}$, van’t Hoff factor is 0.527.
The correct answer is van’t Hoff factor is 0.527.
Note: The value of Van’t Hoff factor obtained is nearing the actual experimental value of i for acetic acid. Van’t Hoff factor is nearly 0.5. [acetic acid has the tendency to dimerize (two molecules combine), so the number of particles after association is 1 and before association is 2. The van’t Hoff factor will be $\dfrac{1}{2}=0.5$]. So, the correct answer can be interpreted easily just by knowing that acetic acid will dimerize. The dimerization of acetic acid looks like
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE