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Standard XII

Chemistry

Resistance and Resistivity

The molar con...

Question

The molar conductivity of $0.04 M$ solution of $M g C l_{2}$ is $200 S c m^{3} m o l^{- 1}$ at $298 K$ . A cell with electrodes that are $2.0 c m^{2}$ in surface area and $0.50 c m$ apart is filled with $M g C l_{2}$ solution. How much current will flow when the potential difference between the two electrodes is $5.0 V$ ?

156.25 A

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0.16 A

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160 A

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None of these

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Solution

The correct option is B $0.16 A$
$Λ_{m} = \frac{κ \times 1000}{M}$

$200 S c m^{3} m o l^{- 1} = \frac{κ \times 1000}{0.04}$

$k = 8 \times 10^{- 3} S$ and $k = C \times cell constant; C = 0.032$ [ $∵$ cell constant $= \frac{l}{A} = \frac{0.5 c m}{2 c m^{2}} = 0.25 c m^{- 1}$ ]

so, resistance $R = 1 / C$

V = 5.0 Volt

From Ohm's law, current $I = \frac{V}{R} = V C = 5 \times 0.032 = 0.16 A$

Hence, the correct option is B.

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The molar conductivity of 0.04 M solution of MgCl2 is 200 Scm3mol−1 at 298 K. A cell with electrodes that are 2.0 cm2 in surface area and 0.50 cm apart is filled with MgCl2 solution. How much current will flow when the potential difference between the two electrodes is 5.0 V?

The correct option is B 0.16 AΛm=κ×1000M200 Scm3mol−1=κ×10000.04k=8×10−3 S and k=C×cell constant;C=0.032 [∵ cell constant =lA=0.5cm2cm2=0.25cm−1]so, resistance R=1/CV = 5.0 VoltFrom Ohm's law, current I=VR=VC=5×0.032=0.16AHence, the correct option is B.