$5.55 g$ of $CaCl X_{2}$ (molar mass $111 g mo l^{- 1}$ ) is dissolved in water to make $1.00 L$ of solution. The conductivity of this solution is $1.02 \times 1 0^{- 2} S c m^{- 1}$ . The equivalent conductivity of the solution, in $S c m^{2} e q^{- 1}$ , is

Answer: A

Molarity

c = 5.55/111 = 0.0500 M

CaCl X_{2}

has n-factor 2, so normality

N = 2 \times 0.0500 = 0.100 N

Λ_{e q} = \frac{1000 κ}{N} = \frac{1000 \times 1.02 \times 1 0 ^{- 2}}{0.100} = 102 S c m^{2} e q^{- 1}

. Using molarity instead gives the molar conductivity

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