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Q.

The limiting molar conductivity of water is equal to

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a

Λm(HCl)0+Λm(NaOH)0Λm(NaCl)0

b

Λm(NaCl)0+Λm(NaOH)0Λm(HCl)0

c

Λm(NaCl)0+Λm(HCl)0Λm(NaOH)0

d

Λm(NaCl)0+Λm(NaOH)0+Λm(HCl)0

answer is D.

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Detailed Solution

Λm(HCl)0+Λm(NaOH)0Λm(NaCl)0=ΛmH+0+ΛmCl0+ΛmNa+0+ΛmOH0ΛmNa+0ΛmCl0
=ΛmH+0+ΛmOH0=ΛmH2O0

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