Question

# At $$25^{\circ}C$$, the molar conductances at infinite dilution for the strong electrolytes $$NaOH, NaCl$$ and $$BaCl_{2}$$ are $$248\times 10^{-4}, 126\times 10^{-4}$$ and $$280\times 10^{-4} Sm^{2} mol^{-1}$$ respectively, $$\lambda_{m}^{\circ}Ba(OH)_{2}$$ in $$Sm^{2} mol^{-1}$$ is

A
52.4×104
B
524×104
C
4.2×104
D
262×104

Solution

## The correct option is D $$524\times 10^{4}$$$$BaCl_{2} + 2NaOH\rightarrow Ba(OH)_{2} + 2NaCl$$$$\lambda^{\infty}_{m\ Ba(OH)_{2}} = \lambda^{\infty}_{m\ BaCl_{2}} + 2\lambda^{\infty}_{m\ NaOH} - 2\lambda^{\infty}_{m\ NaCl}$$$$= 280 \times 10^{-4} + 2\times 248\times 10^{-4} - 2 \times 126\times 10^{-4}$$$$= (280 + 496 - 252) \times 10^{-4}$$$$= 524\times 10^{-4} Sm^{2} mol^{-1}$$. Chemistry

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