At 25-C- the molar conductances at infinite dilution for the strong electrolytes NaOH- NaCl and BaCl2 are 248- 10-4- 126- 10-4 and 280- 10-4Sm2 mol-1 respectively- -m- BaOH2 in S m2 mol-1 is -

The correct option is D 524× 10^ 4 ∧_m^∘Ba(OH)_2 = ∧_m^∘BaCl_2 + 2∧_m^∘NaOH 2∧_m^∘NaCl = 280× 10^ 4 + 2 × 248× 10^ 4 2 × 126× 10^ 4 = 5 ...

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Equivalent, Molar Conductivity and Cell Constant

At 25∘C, th...

Question

At $25^{\circ} C$ , the molar conductances at infinite dilution for the strong electrolytes $N a O H, N a C l$ and $B a C l_{2}$ are $248 \times 10^{- 4}, 126 \times 10^{- 4}$ and $280 \times 10^{- 4} S m^{2} m o l^{- 1}$ respectively. $\land_{m}^{\circ} B a (O H)_{2}$ in $S m^{2} m o l^{- 1}$ is :

52.4 \times 10^{- 4}

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524 \times 10^{- 4}

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402 \times 10^{- 4}

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262 \times 10^{- 4}

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Similar questions

25^{\circ} C

N a O H, N a C l

B a C l_{2}

248 \times 10^{- 4}, 126 \times 10^{- 4}

280 \times 10^{- 4} S m^{2} m o l^{- 1}

λ_{m}^{\circ} B a (O H)_{2}

S m^{2} m o l^{- 1}

B a C l_{2}

10^{- 4}

10^{- 4}

10^{- 4}

m^{2} m o l^{- 1}

\land_{m}^{0}

B a (O H)_{2}

B a C l_{2}

N a C l

N a O H

280 \times 10^{- 4}

126.5 \times 10^{- 4}

248 \times 10^{- 4} S m^{2} m o l^{- 1}

B a (O H)_{2}

B a C l_{2},

248.1 \times 10^{- 4}, 126.5 \times 10^{- 4} a n d 280.0 \times 10^{- 4} S m^{2} m o l^{- 1},

Λ_{m}^{\circ} for B a (O H)_{2}

B a C l_{2}

N a C l

N a O H

280 \times 10^{- 4}

126.5 \times 10^{- 4}

248 \times 10^{- 4} S m^{2} m o l^{- 1}

B a (O H)_{2}

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