When $0.0106 \text{ mole}$ of acetic acid is dissolved in $1 \text{ kg}$ of water, the observed freezing point depression is $0.0205 \text{ K}$. If the calculated freezing point depression is $0.0197 \text{ K}$, the Van't Hoff factor $(i)$ and the degree of dissociation $(\alpha)$ of acetic acid are respectively:
- A$1.041$ and $0.041$
- B$1.041$ and $0.1041$
- C$0.041$ and $0.041$
- D$0.041$ and $1.041$
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Solute $A$ associates in water. When $0.7 \ g$ of solute $A$ is dissolved in $42.0 \ g$ of water,it depresses the freezing point by $0.2^{\circ} C$. The percentage association of solute $A$ in water is $..... \ \%$.
[Given: Molar mass of $A = 93 \ g \ mol^{-1}$. Molal depression constant of water is $1.86 \ K \ kg \ mol^{-1}$]
[Given: Molar mass of $A = 93 \ g \ mol^{-1}$. Molal depression constant of water is $1.86 \ K \ kg \ mol^{-1}$]
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View SolutionWhen $2.44 \ g$ of benzoic acid $(C_6H_5COOH)$ is dissolved in $25 \ g$ of benzene, it shows a depression of freezing point equal to $2.2 \ K$. The molal depression constant of benzene is $5.0 \ K \ kg \ mol^{-1}$. What is the percentage association of the acid, if it forms a dimer in the solution (in $\%$)?
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