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Give A^(@)(1/3 Al^(3+))=63 Omega^(-1) cm...

Give `A^(@)(1/3 Al^(3+))=63 Omega^(-1) cm^(2) mol^(-1)` and `Lambda^(@)(1/2 SO_(4)^(2)) = 80 Omega^(-1)`. The value of `Lambda^(oo)Al_(2)(SO_4)_3` would be

A

`143 Omega^(-1) cm^(2) mol^(-1)`

B

`206 Omega^(-1) cm^(2) mol^(-1)`

C

`286 Omega^(-1) cm^(2) mol^(-1)`

D

`858 Omega^(-1) cm^(2) mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
`Lambda^(@)""_((Al^3+)=Lambda^(@)""_((1/3Al^(3+)))`
`=3 xx 63 =189 Omega^(-1) cm^(2) mol^(-1)Lambda^(@)""_((SO_4^(2-)))=2Lambda^(@)""_((1/2SO_(4)^(2-)))`
`=2 xx 80 =160 Omega^(-1) cm^(2) mol^(-1) =858 Omega^(-1) cm^2 mol^(-1)`
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Given lamda^(oo) [1//2 Mg^(2+)] = 53.06 ohm^(-1) cm^(2) mol^(-1) lamda^(oo) [1//2 SO_(4)^(2-)] = 80 ohm^(-1) cm^(2) mol^(-1) lamda^(oo) [1//3 Al^(3+)] = 63 ohm^(-1) cm^(2) mol^(-1) Calculate the values of Lamda_(m)^(oo) [Al_(2) (SO_(4))_(3)] and Lamda^(oo) [Mg SO_(4)]

lamda_(m)^(0)Na^(+)=150Omega^(-1)cm^(2)"mole"^(-1) : lamda_(eq)^(0)Ba^(2+)=100Omega^(-1)cm^(2)aq^(-1) lamda_(eq)^(0)SO_(4)^(2-)=125Omega^(-1)cm^(2)eq^(-1) : lamda_(m)^(0)Al^(3+)=300Omega^(-1)cm^(2)"mole"^(-1) lamda_(m)^(0)NH_(4)^(+)=200Omega^(-1)cm^(2)"mole"^(-1) : lamda_(m)^(0),Cl^(-)=150Omega^(-1)cm^(2)"mole"^(-1) then calculate (a). lamda_(eq)^(0),Al^(3+) (b). lamda_(eq)^(0)Al_(2)(SO_(4))^(3) (c). lamda_(m)^(0)(NH_(4))NaCl (d). lamda_(0)^(0)NaCl,BaCl_(2).6H_(2)O (e). lamda_(m)^(0),(NH_(4))_(2)SO_(4)Al_(2)(SO_(4))_(3).24H_(2)O ltbr. (f). lamda_(eq)^(0)NaCl

Knowledge Check

  • Given ^^^""^(@)=((1)/(3)Al^(3+))=63 cm^(2)//Omega mol and ^^^ ""^(@)((1)/(2)SO_(4)^(2-))=80 cm^(2)//Omega mol . The value of ^^^ ""^(@) Al_(2)(SO_(4))_(3) would be

    A
    `143 cm^(2)//Omega` mol
    B
    `206 cm^(2)//Omega` mol
    C
    `286 cm^(2)//Omega` mol
    D
    `858 cm^(2)//Omega` mol

  • The conductivity of a saturated solutior of a sparingly soluble salt MX_2 is found to be 4 xx 10^(-5) Omega^(-1) cm^(-1)." If "lambda_(m)^(oo) (1/2 M^(2+))=50 Omega^(-1) cm^(2) mol^(-1) and lambda^(oo) (X^(-))=50 Omega^(-1) cm^(2) mole^(-1) , the solubility product of the salt is about

    A
    `2 xx 10^(-10)M^(3)`
    B
    `2 xx 10^(-12)M`
    C
    `8 xx 10^(-12)M`
    D
    `8 xx 10^(-14)M`

  • The values of ^^_(m)^(oo) for KCl and KNO_(3) are 149.86 and 154.96Omega^(-1)cm^(2)"mol"^(-1) . Also lambda_(Cl)^(oo) is 71.44 ohm^(-1)cm^(2) "mol"^(-1) . The value of lambda_(NO_(3)^(-))^(oo) is

    A
    `76.54 ohm^(-1) cm^(2)"mol"^(-1)`
    B
    `133.08 ohm^(-1)cm^(2)"mol"^(-1)`
    C
    `37.7 ohm^(-1)cm^(2)"mol"^(-1)`
    D
    unpredictable.

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