1.2 , mL of acetic acid is dissolved in water | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $1$. Calculate the mass of acetic acid: $Mass = Density 	imes Volume = 1.02 \, g \, mL^{-1} 	imes 1.2 \, mL = 1.224 \, g$.$2$. Calculate the mol

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) $1$. Calculate the mass of acetic acid: $Mass = Density 	imes Volume = 1.02 \, g \, mL^{-1} 	imes 1.2 \, mL = 1.224 \, g$.$2$. Calculate the moles of acetic acid: $Moles = \frac{1.224 \, g}{60 \, g \, mol^{-1}} = 0.0204 \, mol$.$3$. Calculate molality $(m)$: Since the solution is dilute,we assume the density of the solution is approximately $1 \, g \, mL^{-1}$,so $2 \, L$ of solution is $2 \, kg$ of water. $m = \frac{0.0204 \, mol}{2 \, kg} = 0.0102 \, mol \, kg^{-1}$.$4$. Use the depression in freezing point formula: $\Delta T_{f} = i 	imes K_{f} 	imes m$.$5$. For acetic acid dissociation $(CH_{3}COOH ightleftharpoons CH_{3}COO^{-} + H^{+})$,$i = 1 + \alpha$.$6$. Substitute values: $0.0198 = (1 + \alpha) 	imes 1.85 	imes 0.0102$.$7$. Solve for $\alpha$: $1 + \alpha = \frac{0.0198}{1.85 	imes 0.0102} \approx 1.04928$.$8$. $\alpha = 0.04928$,which is $4.928 \% \approx 5 \%$. The nearest integer is $5$.

Similar Questions

In a solvent,$50\,\%$ of an acid $HA$ dimerizes and the rest dissociates. The van't Hoff factor of the acid is $.....\times 10^{-2}$. (Round off to the nearest integer)

Determine the degree of association (polymerization) for the reaction in aqueous solution $6 HCHO \rightleftharpoons C_6H_{12}O_6$. If the observed (mean) molar mass of the mixture is $150 \ g/mol$. (Given: Molar mass of $HCHO = 30 \ g/mol$)

$75.2 \ g$ of phenol is added to $1 \ kg$ of solvent. The depression in freezing point is $7 \ K$. If phenol undergoes dimerization,calculate the percentage of association. $(K_f = 14 \ K \ kg \ mol^{-1})$

The van't Hoff factor for $BaCl_2$ is $2.47$. Calculate the percentage dissociation of $BaCl_2$ in its aqueous solution. (in $\%$)

If $0.1 \ M$ solution of $NaCl$ is isotonic with $1.1 \ w \%$ urea solution, the degree of ionisation of $NaCl$ is $($Molar masses of urea and $NaCl$ are $60 \ g \ mol^{-1}$ and $58.5 \ g \ mol^{-1}$, respectively.$)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module