$25^{\circ} C$ એ પ્રબળ વિદ્યુતવિભાજ્ય $NaOH ,$ $NaCl$ અને $BaCl _{2}$ માટે અનંત મંદને મોલર વાહક્તા અનુક્રમે $248 \times 10^{-4}, 126 \times 10^{-4}$ અને $280 \times 10^{-4} \,Sm ^{2} mol ^{-1}$ છે. $\lambda_{m}^{o} B a(O H)_{2}$ $S m^{2} m o l^{-1}$ માં $\lambda_{m}^{o} B a(O H)_{2}$ શું હશે ?

Question

A$362 \times 10^{-4}$
B$402 \times 10^{-4}$
C$524 \times 10^{-4}$
D$568 \times 10^{-4}$

AIIMS 2018, Medium

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Solution

c
Given,

$\lambda_{m}^{o}(N a O H)=280 \times 10^{-4} \,Sm ^{2} mol ^{-1}$

$\lambda_{m}^{o}(N a C l)=126 \times 10^{-4} \,Sm ^{2} mol ^{-1}$

$\lambda_{m}^{o}\left(B a C l_{2}\right)=280 \times 10^{-4} \,Sm ^{2} mol ^{-1}$

The reaction for the formation of $B a(O H)_{2}$ can be written as

$B a C l_{2}+2 N a O H \rightarrow B a(O H)_{2}+2 N a C l$

$\because \lambda_{m}^{o} B a(O H)_{2}=\lambda_{m}^{o} B a C l_{2}+\lambda_{m}^{o} N a O H-2 \lambda_{m}^{o} N a C l$

$\therefore \lambda_{m}^{o} B a(O H)_{2}=280 \times 10^{-4}+2 \times 248 \times 10^{-4}-2 \times 126 \times 10^{-4}$

$=(280+496-252) \times 10^{-4}$

$=524 \times 10^{-4} \,Sm ^{2} mol ^{-1}$

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