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Chemistry MCQs for NEET — Practice Questions with Answers

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What is the correct increasing order of molar conductivity at infinite dilution for LiCl, NaC1 and KCI ?

A. $ KCl \lt NaCl \lt LiCl $ B. $ LiCl \lt NaCl \lt KCl $ C. $ LiCl \lt KCl \lt NaCl $ D. $ NaCl \lt KCl \lt LiCl $

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Explanation

The molar conductivity at infinite dilution follows the order: $LiCl < NaCl < KCl$. This is because the ionic mobility increases with decreasing ionic radius and charge density. The smaller $Li^+$ ion has a higher charge density and gets more strongly solvated, reducing its mobility.

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Resistance and specific conductance Of the cell having N/50 KCI solution is $ 400\Omega and 0.002765 S. cm^{-1} $ respectively then what is the cell constant ofthe cell?

A. $ 6.19 cm ^ {-1} $ B. $ 1.106 cm ^ {-1} $ C. $ 14.46 cm ^ {-1} $ D. $ 2.212 cm ^ {-1} $

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Explanation

Cell constant (K) is defined as the ratio of the distance between the electrodes (l) to the area of the electrodes (A). K = l/A. Given resistance R = 400 Î© and specific conductance Îº = 0.002765 S.cmâ»Â¹, we can calculate K using the equation R = (1/ÎºK). Substituting the values, we get K = 1.106 cmâ»Â¹.

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At 298k temperature resistance of the 0.05M solution is $ 3.16 \Omega $ . What will be the equivalent eonduetance of the same if cell consmnt is $ 0367 cm ^ {-1} $ ?

A. $ 0.2322 S . cm^2. g equivalent ^ {-1} $ B. $ 2.322 S . cm^2. g equivalent ^ {-1} $ C. $ 23.22 S . cm^2. g equivalent ^ {-1} $ D. $ 232.28 S . cm^2. g equivalent ^ {-1} $

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Resistance of the 0.05M solution placed between two Pt electrodes havning surface area $ 10 cm^ 2 at 1.5cm distance is 50 \Omega $ , what is its molar conductance?

A. $ 66.6 S.cm ^2 .mol ^ {-1} $ B. $ 15 S.cm ^2 .mol ^ {-1} $ C. $ 60 S.cm ^2 .mol ^ {-1} $ D. $ 6.66 S.cm ^2 .mol ^ {-1} $

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$ \lambda ^ \circ _{ClCH_2 COONa} = 2.24 S.m^2 mol ^ {-1} , \lambda ^ \circ _{NaCl} = 38.2 S. m^2 .mol ^ {-1} and |lambda ^ \circ _{HCl} = 203 S.m^2 .mol ^ {-1} , then what is the value of \lambda ^ \circ _{ ClCH_2 COOH} $ ?

A. $ 288.5 S . cm^2. g equivalent ^ {-1} $ B. $ 289.5 S . cm^2. g equivalent ^ {-1} $ C. $ 388.8 S . cm^2. g equivalent ^ {-1} $ D. $ 59.5 S . cm^2. g equivalent ^ {-1} $

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On applying potential difference or 20V for 2 minute to the electric wire having$ 10 \Omega $ how much quantity of electricity will passed through it?

A. 120 C B. 240C C. 20C D. 4C

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Specific conductance of two solution A and B found to be $ K_1 and K_2 $ respectiveluy on measuring by the same conductivity cell. if same volume of both solution are taken in a conductivity cell having cell constant x then what will be the resistance of the mixture? ( consider that there is no difference in the degree Of dissociatioin Of solutions on mixing)

A. $ R = { K_1 + K_2 \over 2x} $ B. $ R = { 2 (K_1 +K_2 ) \over x } $ C. $ 2x ( K_1 +K_2 ) $ D. $ R = { 2x \over K_1 + K_2 } $

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Explanation

The resistance (R) of a solution is inversely proportional to its specific conductance (K). If two solutions with specific conductances Kâ‚ and Kâ‚‚ are mixed in equal volumes, the specific conductance of the mixture (K) will be the arithmetic mean of Kâ‚ and Kâ‚‚, assuming no change in dissociation. Thus, K = (Kâ‚ + Kâ‚‚)/2. Since R âˆ 1/K, the resistance of the mixture will be R = constant/(Kâ‚ + Kâ‚‚)/2 = 2 Ã— constant/(Kâ‚ + Kâ‚‚), where the constant is related to the cell constant.

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What is the molar conductiviy at infinite dilution of the solution having concentration 0.01M and molar conductivity is $ 19.6 S. cm^2 mol ^{-1} ? at 298K temp dissociation constant of weak electrolyte is 2.5 \times 10 ^{-5} $ .

A. $ 250 S .cm ^2 . mol ^{-1} $ B. $ 196 S.cm^2 .mol ^ {-1} $ C. $ 392 S . Cm ^2 . Mol ^{-1} $ D. $ 384 S .cm^2 mol ^ {-1} $

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At 300 K temperature svwcific conductance and limiting conductance of the 0.7 M aqueous solution of acetic acid are $ 0.01365 S . Cm ^{-1} and 390 S. cm^2 . Mol^{-1} $
respectively then what is the dissociation constant Of the acetic acid?

A. $ 1.75 \times 10^ {-6} M $ B. $ 1.75 \times 10^ {-5 } M $ C. $ 3.5 \times 10^ {-4} M $ D. $ 3.5 \times 10^ {-6} M $

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If concentration of the $ Cu ^{2+} _{(aq)} is made double then what change is observed in the E ^ 0 _cell Cu_{(s)} + 2Ag^{2+}_{(Ã¤q)} \rightleftharpoons Cu^{2+}_{(aq)}+ 2Ag_{(s)} E ^\circ = 0.48 Volt $ for the reaction

A. doubled B. halve C. no change D. little less than double

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Explanation

The Nernst equation for the given cell is: E_cell = E^0_cell - (RT/nF) ln(Q), where Q is the reaction quotient. Doubling the concentration of Cu^{2+}_{(aq)} does not affect the standard cell potential (E^0_cell), as it depends only on the standard reduction potentials of the half-reactions involved. Therefore, the cell potential (E_cell) remains unchanged.

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