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_3COONa ` and NaCl respectively ?A. 97.65 , 111.0 and 242.8 S `"cm"^(2) "mol"^(-1)`B. 195.3, 182.0 and 26.2 S `"cm"^(2) "mol"^(-1)`C. 271.6, 91.0 and 126.4 S `"cm"^(2) "mol"^(-1)`D. 119.0, 1024.5 and 9.2 S `"cm"^(2) "mol"^(-1)`

Answer» Correct Answer - C
`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

Related InterviewSolutions