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Standard XII

Chemistry

Variation of Specific, Equivalent and Molar Conductivities with Concentration

Calculate mol...

Question

Calculate molar conductivity for $N H_{4} O H (S c m^{2} m o l^{- 1})$ , given that molar conductivity for $B a (O H)_{2}, B a C l_{2}$ and $N H_{4} C l$ for 523.28, 280.0 and 129.8 $S c m^{2} m o l^{- 1}$ respectively.

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Solution

The molar conductivity of $N H_{4} O H$ can be calculated as
$λ_{N H_{4} O H}^{0} = \frac{λ_{B a (O H)_{2}}^{0} + 2 λ_{N H_{4} C l}^{0} - λ_{B a C l_{2}}^{0}}{2} = \frac{523.28 + 2 \times 129.8 - 280}{2} = 251.44 S c m^{2} m o l^{- 1}$

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Calculate molar conductivity for NH4OH(Scm2mol–1), given that molar conductivity for Ba(OH)2,BaCl2 and NH4Cl for 523.28, 280.0 and 129.8 Scm2 mol–1 respectively.

The molar conductivity of NH4OH can be calculated as λ0NH4OH=λ0Ba(OH)2+2λ0NH4Cl−λ0BaCl22=523.28+2×129.8−2802=251.44 Scm2mol−1

Calculate molar conductivity for $N H_{4} O H (S c m^{2} m o l^{- 1})$ , given that molar conductivity for $B a (O H)_{2}, B a C l_{2}$ and $N H_{4} C l$ for 523.28, 280.0 and 129.8 $S c m^{2} m o l^{- 1}$ respectively.

The molar conductivity of $N H_{4} O H$ can be calculated as
$λ_{N H_{4} O H}^{0} = \frac{λ_{B a (O H)_{2}}^{0} + 2 λ_{N H_{4} C l}^{0} - λ_{B a C l_{2}}^{0}}{2} = \frac{523.28 + 2 \times 129.8 - 280}{2} = 251.44 S c m^{2} m o l^{- 1}$