A solution containing $2.08$ g of anhydrous barium chloride is $400$ CC of water has a specific conductivity $0.0058$ $o h m^{- 1}$ $c m^{- 1}$ . The molar conductivities of this solution is 131.5 $o h m^{- 1}$ $c m^{- 1}$ .

True

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False

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Solution

The correct option is B False
Mass of $B a {C l}_{2}$ =2.08g
Mass of Solution=400 ${c m}^{3}$ =400ml
Conductivity(k)=0.0058 ${o h m}^{- 1}$ ${c m}^{- 1}$
Molarity(m)= $\frac{M o l e s o f B a {C l}_{2}}{V o l u m e (i n L)}$
= $\frac{2.08 g / 208 g {m o l}^{- 1}}{0.4}$
=0.025M.
Molar conductivity= $\frac{k \times 1000}{M}$
= $\frac{0.0058 \times 1000}{0.025}$
Molar conductivity=232 ${o h m}^{- 1} {c m}^{- 1}$ .
Therefore it is a false statement.

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