1.

What is true for Lamda_(m(NH_(4)OH))^(@) ?

Answer»

`Lamda_(m(NaOH))^(@)+Lamda_(m(NH_(4)Cl))^(@)-Lamda_(m(HCl))^(@)`
`Lamda_(m(NH_(4)Cl))^(@)+Lamda_(m(NaOH))^(@)-Lamda_(m(NACL))^(@)`
`Lamda_(m(NH_(4)Cl))^(@)+Lamda_(m(NaCl))^(@)-Lamda_(m(NaOH))^(@)`
`Lamda_(m(NH_(4)Cl))^(@)+Lamda_(m(NH_(4)Cl))^(@)-Lamda_(m(NaOH))^(@)`

Solution :(i) `Lamda_(m(NH_(4)Cl))^(@)=lamda_(m(NH_(4)^(+)))^(@)+lamda_(m(Cl^(-)))^(@)`
(ii) `Lamda_(m(NaOH))^(@)=lamda_(m(Na^(+)))^(@)+lamda_(m(OH^(-)))^(@)`
(iii) `Lamda_(m(NaCl))^(@)=lamda_(m(Na^(+)))^(@)+lamda_(m(Cl^(-)))^(@)`
`therefore (i)+(ii)-(iii)=lamda_(m(NH_(4)^(+)))^(@)+lamda_(m(OH^(-)))^(@)`
`=lamda_(m(NH_(4)Cl))^(@)`, so OPTION (B) is right.
So, if REPEAT accordingly, then option (A), (C) and (D) are wrong, as they are not the value of `Lamda_(m)^(@)(NH_(4)OH)`.


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