ElectrochemistryHard
Question
A big irregular shaped vessel contained water, conductivity of which was 2.56 × 10−3 S−1m−1. 585 g of NaCl was then added to the water and conductivity after the addition of NaCl, was found to be 3.06 × 10−3 S−1m−1. The molar conductivity of NaCl at this concentration is 1.5 × 10−2 S−1m2 mol−1. The capacity of vessel if it is fulfilled with water, is
Options
A.3 × 104 L
B.30 L
C.3 × 108 L
D.3 × 105 L
Solution
$\Lambda_{m} = \frac{K}{C} \Rightarrow 1.5 \times 10^{- 2} = \frac{3.06 \times 10^{- 3} - 2.56 \times 10^{- 3}}{C}$
$\therefore C = \frac{1}{30}\text{mol}m^{- 3} = \frac{1}{30} \times 10^{- 3}\text{ mol.}\text{l}^{- 1} = \frac{585/58.5}{V} $$$\therefore V = 3 \times 10^{5}\text{ L}$$
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