back to directory
NEET CHEMISTRYElectrochemistryMedium

Question

The molar conductivity of 0.007 M0.007\text{ M} acetic acid is 20 S cm2 mol120\text{ S cm}^2\text{ mol}^{-1}. The dissociation constant of acetic acid is: (ΛH+o=350 S cm2 mol1\Lambda^o_{H^+} = 350\text{ S cm}^2\text{ mol}^{-1} and ΛCH3COOo=50 S cm2 mol1\Lambda^o_{CH_3COO^-} = 50\text{ S cm}^2\text{ mol}^{-1})

A

1.75×105 mol L11.75 \times 10^{-5}\text{ mol L}^{-1}

B

2.50×105 mol L12.50 \times 10^{-5}\text{ mol L}^{-1}

C

1.75×104 mol L11.75 \times 10^{-4}\text{ mol L}^{-1}

D

2.50×104 mol L12.50 \times 10^{-4}\text{ mol L}^{-1}

Step-by-Step Solution

Given: Concentration, c=0.007 Mc = 0.007\text{ M} Molar conductivity, Λm=20 S cm2 mol1\Lambda_m = 20\text{ S cm}^2\text{ mol}^{-1} Limiting molar conductivity of H+H^+, ΛH+=350 S cm2 mol1\Lambda^{\circ}_{H^+} = 350\text{ S cm}^2\text{ mol}^{-1} Limiting molar conductivity of CH3COOCH_3COO^-, ΛCH3COO=50 S cm2 mol1\Lambda^{\circ}_{CH_3COO^-} = 50\text{ S cm}^2\text{ mol}^{-1}

According to Kohlrausch's law, limiting molar conductivity of acetic acid: Λm(CH3COOH)=ΛH++ΛCH3COO\Lambda^{\circ}_m(CH_3COOH) = \Lambda^{\circ}_{H^+} + \Lambda^{\circ}_{CH_3COO^-} Λm=350+50=400 S cm2 mol1\Lambda^{\circ}_m = 350 + 50 = 400\text{ S cm}^2\text{ mol}^{-1}

Degree of dissociation, α=ΛmΛm=20400=0.05\alpha = \frac{\Lambda_m}{\Lambda^{\circ}_m} = \frac{20}{400} = 0.05

Dissociation constant, Ka=cα21αK_a = \frac{c\alpha^2}{1 - \alpha} Assuming 1α11 - \alpha \approx 1 (since α\alpha is small): Kacα2=0.007×(0.05)2=0.007×0.0025=1.75×105 mol L1K_a \approx c\alpha^2 = 0.007 \times (0.05)^2 = 0.007 \times 0.0025 = 1.75 \times 10^{-5}\text{ mol L}^{-1}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Electrochemistry. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYElectrochemistryconductivityaceticcmtextdissociationconstant

More Electrochemistry Questions

View all

A cell reaction becomes spontaneous when:

A.$\Delta G^{\circ}$ is negative
B.$\Delta G^{\circ}$ is positive
C.$E^{\circ}_{Red}$ is positive
D.$E^{\circ}_{Red}$ is negative
EasySolve

In a typical fuel cell, the reactants (R) and products (P) are:

A.R = $H_2(g)$, $O_2(g)$; P = $H_2O_2(l)$
B.R = $H_2(g)$, $O_2(g)$; P = $H_2O(l)$
C.R = $H_2(g)$, $O_2(g)$, $Cl_2(g)$; P = $HClO_4(aq)$
D.R = $H_2(g)$, $N_2(g)$; P = $NH_3(aq)$
EasySolve

The cell reaction of an electrochemical cell is $Cu^{2+}(C_1) + Zn \rightarrow Cu + Zn^{2+}(C_2)$. The change in free energy will be the function of:

A.$\ln(C_1+C_2)$
B.$\ln(C_2/C_1)$
C.$\ln C_2$
D.$\ln C_1$
MediumSolve

The efficiency of a fuel cell is given by:

A.$\Delta H/\Delta G$
B.$\Delta G/\Delta S$
C.$\Delta G/\Delta H$
D.$\Delta S/\Delta G$
EasySolve

The number of Faradays (F) required to produce $20 \text{ g}$ of calcium from molten $\text{CaCl}_2$ (Atomic mass of Ca = $40 \text{ g mol}^{-1}$) is:

A.2
B.3
C.4
D.1
MediumSolve

For a cell involving one electron $E^\ominus_{\text{cell}} = 0.59 \text{ V}$ at $298 \text{ K}$. The equilibrium constant for the cell reaction is: [Given that $\frac{2.303 RT}{F} = 0.059 \text{ V}$ at $T = 298 \text{ K}$]

A.$1.0 \times 10^{30}$
B.$1.0 \times 10^2$
C.$1.0 \times 10^5$
D.$1.0 \times 10^{10}$
EasySolve

The value of $E^{\circ}_{cell}$ for the following reaction is: $Cu^{2+} + Sn^{2+} \rightarrow Cu + Sn^{4+}$ (Given, equilibrium constant is $10^6$)

A.0.17
B.0.01
C.0.05
D.1.77
MediumSolve

The three cells with their $E^{\circ}_{cell}$ values are given below: 1. $Fe|Fe^{2+}||Fe^{3+}|Fe \ ; \ 0.404\text{ V}$ 2. $Fe|Fe^{2+}||Fe^{3+}, Fe^{2+}|Pt \ ; \ 1.211\text{ V}$ 3. $Fe|Fe^{3+}||Fe^{3+}, Fe^{2+}|Pt \ ; \ 0.807\text{ V}$ The standard Gibbs free energy change values for three cells are, respectively: (F represents the charge on $1\text{ mole}$ of electrons.)

A.–1.212 F, –1.211 F, –0.807 F
B.+2.424 F, +2.422 F, +2.421 F
C.–0.808 F, –2.422 F, –2.421 F
D.–2.424 F, –2.422 F, –2.421 F
HardSolve

This neet chemistry practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice chemistry sets →