The molar conductance's at infinite dilution of BaCl2, NaCl and NaOH are respectively 280×10−4,126.5×10−4,248×10−4Sm2mol−1The molar conductance at infinite dilution for Ba(OH)2 is

μBaCl2∞=μBa2+∞+2μC1−∞...(i)μNaC1∞=μNa+∞+μCl−∞....(ii)μNaOH∞=μNa+∞+μHO−∞.....(iii)for Ba(OH)2(i)+2( iii )−2( ii )⇒μBa(OH)2=280×10−4+2248×10−4−2126.5×10−4 =523×10−4Sm2mol−1

The molar conductance's at infinite dilution of BaCl2, NaCl and NaOH are respectively 280×10−4,126.5×10−4,248×10−4Sm2mol−1The molar conductance at infinite dilution for Ba(OH)2 is

μBaCl2∞=μBa2+∞+2μC1−∞...(i)μNaC1∞=μNa+∞+μCl−∞....(ii)μNaOH∞=μNa+∞+μHO−∞.....(iii)for Ba(OH)2(i)+2( iii )−2( ii )⇒μBa(OH)2=280×10−4+2248×10−4−2126.5×10−4 =523×10−4Sm2mol−1

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The molar conductance's at infinite dilution of BaCl₂, NaCl and NaOH are respectively
$280 \times 10^{- 4}, 126 .5 \times 10^{- 4}, 248 \times 10^{- 4} {S m}^{2} {m o l}^{- 1}$ The molar conductance at infinite dilution for Ba(OH)₂ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$523 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$

$65 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$

$52 .3 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$

$5.23 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$μ_{{BaCl}_{2}}^{\infty} = μ_{{Ba}^{2 +}}^{\infty} + 2 μ_{C 1^{-}}^{\infty}$ ...(i) $μ_{NaC 1}^{\infty} = μ_{{Na}^{+}}^{\infty} + μ_{{Cl}^{-}}^{\infty}$ ....(ii) $μ_{NaOH}^{\infty} = μ_{{Na}^{+}}^{\infty} + μ_{{HO}^{-}}^{\infty}$ .....(iii)for $Ba (OH)_{2} (i) + 2 (iii) - 2 (ii) \Rightarrow μ_{(Ba (OH)_{2})}$ $= (280 \times 10^{- 4}) + 2 (248 \times 10^{- 4}) - 2 (126 .5 \times 10^{- 4})$ $= 523 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$