If the specific conductance of 1M H2SO4 solution is 26 - 10-2 S cm-1- then the equivalent conductivity would be-

The specific conductance of saturated solution of AgCl is found to be $1.86 \times 10^{- 6} o h m^{- 1} c m^{- 1}$ and that of water is $6 \times 10^{- 8} o h m^{- 1} c m^{- 1}$ . The solubility of AgCl given that $\land_{A g C l}^{\circ}$ = $137.2 o h m^{- 1} c m^{2} e q^{- 1}$ is

Q. Calculate the equivalent conductivity of

1 M H_{2} S O_{4}

solution if its conductivity is

26 \times 10^{- 2} o h m^{- 1} c m^{- 1}

(Atomic weight of sulphur = 32 g / m o l)

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If the specific conductance of 1M H2SO4 solution is 26×10−2 S cm−1, then the equivalent conductivity would be:

The correct option is B 1.3×102Scm2eq.−1Normality=nf×Molarity =2×1=2NNow,λeq=1000×kN⇒λeq=1000×26×10−22=1.3×102Scm2eq.−1

If the specific conductance of $1 M H_{2} S O_{4}$ solution is $26 \times 10^{- 2}$ S $c m^{- 1}$ , then the equivalent conductivity would be:

The correct option is B $1.3 \times 10^{2} S c m^{2} e q .^{- 1}$
$N o r m a l i t y = n_{f} \times M o l a r i t y$
$= 2 \times 1 = 2 N$
Now,
$λ e q = 1000 \times \frac{k}{N}$
$\Rightarrow λ_{e q} = \frac{1000 \times 26 \times 10^{- 2}}{2} = 1.3 \times 10^{2} S c m^{2} e q .^{- 1}$