1.

Limiting molar conductivity for some ions is given below (in S cm^(2) mol^(-1) ) : Na^+ -50.1 , Cl^(-) - 76.3 , H^(+) - 349.6 , CH_3COO^(-) -40.9 , Ca^(2+) -119.0 What will be the limiting molar conductivities (Lambda_m^@) of CaCl_2, CH_3COONaand NaCl respectively ?

Answer»

97.65 , 111.0 and 242.8 S `"cm"^(2) "MOL"^(-1)`
195.3, 182.0 and 26.2S `"cm"^(2) "mol"^(-1)`
271.6, 91.0 and 126.4S `"cm"^(2) "mol"^(-1)`
119.0, 1024.5 and 9.2S `"cm"^(2) "mol"^(-1)`

Solution :`Lambda_(m" " CaCl_2)^@=lambda_(Ca^(2+))^@ + 2lambda_(CL^-)^@`
=119.0 + 2 x 76.3 = 271.6 S `cm^2 mol^(-1)`
`Lambda_(m " " CH_3COONa)^@=Lambda_(CH_3COO^-)^@+lambda_(NA^+)^@`
=40.9+50.1=91 S `cm^2 mol^(-1)`
`Lambda_(m " " NACL)^@=lambda_(Na^+)^@ + lambda_(Cl^-)^@`
= 50.1 + 76.3 = 126.4 S `cm^2 mol^(-1)`


Discussion

No Comment Found