
Electrolyte $\text{KCl}$ $\text{KN}{{\text{O}}_{3}}$ $\text{HCl}$ $\text{NaOAc}$ $\text{NaCl}$ ${}_{\text{(Sc}{{\text{m}}^{2}}}{{\wedge }^{\infty }}{}_{\text{mo}{{\text{l}}^{-1}})}$ 149.9 145.0 426.2 91.0 126.5
Calculate ${{\wedge }^{\infty }}_{\text{HOAc}}$ using appropriate molar conductance of the electrolytes listed above at infinite dilution in ${{\text{H}}_{2}}\text{O}$ at $25{}^\circ \text{C}$.
A. 517.2
B. 552.7
C. 390.7N
D. 217.5
| Electrolyte | $\text{KCl}$ | $\text{KN}{{\text{O}}_{3}}$ | $\text{HCl}$ | $\text{NaOAc}$ | $\text{NaCl}$ |
| ${}_{\text{(Sc}{{\text{m}}^{2}}}{{\wedge }^{\infty }}{}_{\text{mo}{{\text{l}}^{-1}})}$ | 149.9 | 145.0 | 426.2 | 91.0 | 126.5 |
Answer
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Hint: The ratio of conductivity of an electrolyte and its molar conductance gives the molar conductance, which is denoted by ${{\wedge }_{\text{m}}}\text{ }$. The molar conductance has SI unit, Siemens metre squared per mol $\text{(S}{{\text{m}}^{2}}\text{mo}{{\text{l}}^{-1}})$.
Complete Answer:
-We have to calculate the molar conductance at infinite dilution for $\text{HOAc}$ at a given temperature.
-so, firstly we will write the reaction between hydrochloric acid and $\text{NaOAc}$ i.e.
$\text{HCl + NaOAc }\to \text{ NaCl + HOAc}$
-Now, for the calculating the molar conductance the formula will be written according to the reaction but the temperature will not
-Now, the formula of molar conductance at infinite dilution will be:
$\wedge _{\text{HOAc}}^{\infty }\text{ = }\wedge _{\text{NaOAc}}^{\infty }\text{ + }\wedge _{\text{HCl}}^{\infty }\text{ - }\wedge _{\text{NaCl}}^{\infty }\text{ }....\text{(1)}$
-Here, the molar conductance of sodium chloride is subtracted because molar conductance of sodium chloride is not asked i.e. to remove the sodium from sodium acetate and chlorine from hydrochloric acid.
-It is given in the question that the value of molar conductance of sodium acetate, hydrochloric acid and sodium chloride is 91.0, 426.2 and 126.5 respectively.
-Now, we will put these values in the equation (1) and we will get,
\[91\text{ + 426}\text{.2 - 126}\text{.5 = 390}\text{.7 Sc}{{\text{m}}^{2}}\text{mo}{{\text{l}}^{-1}}\].
Therefore, option C is the correct answer.
Note: According to the theory of electrolysis, dissociation which was given by Arrhenius, if the electrolyte has completely dissociated then the value of molar conductance at infinite dilution will be minimum.
Complete Answer:
-We have to calculate the molar conductance at infinite dilution for $\text{HOAc}$ at a given temperature.
-so, firstly we will write the reaction between hydrochloric acid and $\text{NaOAc}$ i.e.
$\text{HCl + NaOAc }\to \text{ NaCl + HOAc}$
-Now, for the calculating the molar conductance the formula will be written according to the reaction but the temperature will not
-Now, the formula of molar conductance at infinite dilution will be:
$\wedge _{\text{HOAc}}^{\infty }\text{ = }\wedge _{\text{NaOAc}}^{\infty }\text{ + }\wedge _{\text{HCl}}^{\infty }\text{ - }\wedge _{\text{NaCl}}^{\infty }\text{ }....\text{(1)}$
-Here, the molar conductance of sodium chloride is subtracted because molar conductance of sodium chloride is not asked i.e. to remove the sodium from sodium acetate and chlorine from hydrochloric acid.
-It is given in the question that the value of molar conductance of sodium acetate, hydrochloric acid and sodium chloride is 91.0, 426.2 and 126.5 respectively.
-Now, we will put these values in the equation (1) and we will get,
\[91\text{ + 426}\text{.2 - 126}\text{.5 = 390}\text{.7 Sc}{{\text{m}}^{2}}\text{mo}{{\text{l}}^{-1}}\].
Therefore, option C is the correct answer.
Note: According to the theory of electrolysis, dissociation which was given by Arrhenius, if the electrolyte has completely dissociated then the value of molar conductance at infinite dilution will be minimum.
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