If the molar conductance values of Ca2- and Cl- at infinite dilution are respectively 118-88 - 10-4cm2 mho mol-1 and 77-66 - 10-4cm2 mho mol-1- then that of CaCl2 is cm2 mho mol-1

Λ_m ^∘(CaCl_2)=Λ_m ^∘(Ca^2+)+2Λ_m ^∘(Cl^ ) =118.88 × 10^ 4+2(77.33 × 10^ 4) =273.54 × 10^ 4 cm^2 mho mol^ 1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Chemistry

Electrolysis and Electrolytes

If the molar ...

Question

If the molar conductance values of $C a^{2 +}$ and $C l^{-}$ at infinite dilution are respectively $118.88 \times 10^{- 4} c m^{2} m h o m o l^{- 1}$ and $77.66 \times 10^{- 4} c m^{2} m h o m o l^{- 1}$ , then that of $C a C l_{2}$ is ( $c m^{2} m h o m o l^{- 1}$ )

118.88 \times 10^{- 4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

196.21 \times 10^{- 4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

154.66 \times 10^{- 4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

273.54 \times 10^{- 4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $273.54 \times 10^{- 4}$
$Λ_{m}^{\circ} (C a C l_{2}) = Λ_{m}^{\circ} (C a^{2 +}) + 2 Λ_{m}^{\circ} (C l^{-})$
$= 118.88 \times 10^{- 4} + 2 (77.33 \times 10^{- 4})$
$= 273.54 \times 10^{- 4} c m^{2} m h o m o l^{- 1}$

Suggest Corrections

Similar questions

Q. If the molar conductance values of

C a^{2 +}

and

C l^{-}

at infinite dilution are respectively

118.88 \times 10^{- 4} m^{2} m h o m o l^{- 1}

and

77.33 \times 10^{- 4} m^{2} m h o m h o^{- 1}

then that of

C a C l_{2}

is:

Q. If the molar conductance values of

C a^{2 +}

and

C l^{-}

at infinite dilution are respectively

118.88 \times 10^{- 4} m^{2} {mho mol}^{- 1}

and

77.33 \times 10^{- 4} m^{2} {mho mol}^{- 1}

, then that of

C a C l_{2}

(in m^{2} {mho mol}^{- 1})

is:

Q. The molar conductivities at infinite dilution of

K C l, K N O_{3}

and

A g N O_{3}

298 K

are

0.01499 m h o m^{2} m o l^{- 1}, 0.01250 m h o m^{2} m o l^{- 1}

and

0.01334 m h o m^{2} m o l^{- 1}

respectively. What is the molar conductivity of

A g C l

infinite dilution at this temperature?

Q. The molar conductivities of

N H_{4}^{+}

ion and

C l^{-}

ion are

73.5 m h o c m^{2} m o l^{- 1}

and

76.2 m h o c m^{2} m o l^{- 1}

respectively. The specific conductivity of

0.1 M N H_{4} C l

1.22 \times 10^{- 2} m h o c m^{- 1}

. What is the dissociation constant of

N H_{4} C l

Q. The specific conductivity of a saturated solution of silver chloride is

2.30 \times 10^{- 6} m h o c m^{- 1}

25^{\circ} C

. Calculate the solubility of silver chloride at

25^{\circ} C

λ_{A g^{+}} = 61.9 m h o c m^{2} m o l^{- 1}

and

λ_{C l^{-}} = 76.3 m h o c m^{2} m o l^{- 1}

Join BYJU'S Learning Program

If the molar conductance values of Ca2+ and Cl− at infinite dilution are respectively 118.88×10−4cm2 mho mol−1 and 77.66×10−4cm2 mho mol−1, then that of CaCl2 is (cm2 mho mol−1)

The correct option is D 273.54×10−4Λ∘m(CaCl2)=Λ∘m(Ca2+)+2Λ∘m(Cl−) =118.88×10−4+2(77.33×10−4) =273.54×10−4 cm2 mho mol−1

If the molar conductance values of $C a^{2 +}$ and $C l^{-}$ at infinite dilution are respectively $118.88 \times 10^{- 4} c m^{2} m h o m o l^{- 1}$ and $77.66 \times 10^{- 4} c m^{2} m h o m o l^{- 1}$ , then that of $C a C l_{2}$ is ( $c m^{2} m h o m o l^{- 1}$ )

The correct option is D $273.54 \times 10^{- 4}$
$Λ_{m}^{\circ} (C a C l_{2}) = Λ_{m}^{\circ} (C a^{2 +}) + 2 Λ_{m}^{\circ} (C l^{-})$
$= 118.88 \times 10^{- 4} + 2 (77.33 \times 10^{- 4})$
$= 273.54 \times 10^{- 4} c m^{2} m h o m o l^{- 1}$