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Question: For \(Na^+\) , the value of the symbol \({{\lambda}_m}^0\) is \(50.0{{\Omega }^{-1}}{{cm}^{2}}mo{{l}...

For Na+Na^+ , the value of the symbol λm0{{\lambda}_m}^0 is 50.0Ω−1cm2mol−150.0{{\Omega }^{-1}}{{cm}^{2}}mo{{l}^{-1}} . The speed of Na+Na^+ ion in the solution, if in the cell, electrodes are 5  cm5\;cm apart and to which a potential of 19.3  volt19.3\;volt is applied is
A. 2×10−3cm  s−12\times {{10}^{-3}}cm\;{{s}^{-1}}
B. 1×10−3cm  s−11\times {{10}^{-3}}cm\;{{s}^{-1}}
C. 2×10−4cm  s−12\times {{10}^{-4}}cm\;{{s}^{-1}}
D. 2×10−2cm  s−12\times {{10}^{-2}}cm\;{{s}^{-1}}

Explanation

Solution

The value given here for the symbol λm0{{\lambda}_m}^0 is known as the molar ionic conductivity. Using this value we can find the value of the ionic mobility, which can be defined as the molar conductivity of the solution per charge flowing by one mole of electrons (also known as Faraday). The ionic mobility can also be defined as the drift velocity per unit electric field. From this, the value of the velocity of ions can be obtained.

Complete step by step answer:
Let us note down the given data and data we need to find;
Molar conductivity of Na+Na^+ ion 50.0Ω−1cm2mol−150.0{{\Omega }^{-1}}{{cm}^{2}}mo{{l}^{-1}}
Distance between the electrodes d=5cmd=5cm
Potential difference applied between two ends V=19.3voltV=19.3volt
Mobility of Na+Na^+ ion μm0=?{{\mu }_{m}}^{0}=?
Velocity or speed of Na+Na^+ ion v=?v=?

The molar conductivity can be defined as conductivity of the ion in the solution with volume VV . The ionic mobility is found to be proportional to the molar conductivity of the solution, and the proportionality constant is the charge of one mole of electrons (also known as Faraday) which is mathematically expressed as
λm0=Fμm0{{\lambda }_{m}}^{0}=F{{\mu }_{m}}^{0}
From the above equation, mobility is expressed as
μm0=λm0F{{\mu }_{m}}^{0}=\dfrac{{{\lambda }_{m}}^{0}}{F}
Substituting the value of molar conductivity and Faraday=96500  CoulombFaraday=96500\;Coulomb
μm0=50ohm−1cm2mol−196500Asec⁡{{\mu }_{m}}^{0}=\dfrac{50oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}}{96500A\sec }
For the units, from Ohm’s Law, ΩA=Volt\Omega A=Volt
μm0=5.18×10−4volt−1cm2sec⁡−1{{\mu }_{m}}^{0}=5.18\times {{10}^{-4}}vol{{t}^{-1}}c{{m}^{2}}{{\sec }^{-1}}

The ionic mobility can be defined as the velocity achieved by an ion moving through a gas under an electric field. Hence, the ionic mobility can be mathematically expressed as
μm0=ionic velocity (cm sec−1)Potential difference (volt)Distance between electrodes (cm){{\mu }_{m}}^{0}=\dfrac{\text{ionic velocity (cm se}{{\text{c}}^{-1}})}{\dfrac{\text{Potential difference (volt)}}{\text{Distance between electrodes (cm)}}}
From the above formula, the ionic velocity is expressed as
v=μm0×Vdv={{\mu }_{m}}^{0}\times \dfrac{V}{d}
Substituting the given values,
v=5.18×10−4×19.35v=5.18\times {{10}^{-4}}\times \dfrac{19.3}{5}
∴v=2×10−3cm  s−1\therefore v=2\times {{10}^{-3}}cm\;{{s}^{-1}}

Hence, the correct answer is option A.

Note: Here, as the name of the symbol λm0{{\lambda}_m}^0 is not given, we should be able to remember it as the molar conductivity of the solution, which is defined as the conductivity of volume VV having 1  mole1\;mole solution. The unit of faraday is A  secA\;sec or coulomb  coulomb\; but, it shows the charge of 1  mole1\;mole of electrons. Hence, the unit of   mol\;mol is present but not written for Faraday and due to this, the unit of   mol\;mol is common from the numerator and denominator, and thus canceled.