`8.0575xx10^(-2) kg` of Glauber's slat is dissolved in water to obtain `1 dm^(3)` of a solution of density `1077.2 kg m^(-3)`. Calculate the molarity, molality and mole fraction of `Na_(2)SO_(4)` in solution.

Answer

(i) 0.25 M (ii) 0.251 (iii) `chi_(Na_(2)SO_(4)) = 0.0045`

Explanation

(i) 0.25 M (ii) 0.251 (iii) `chi_(Na_(2)SO_(4)) = 0.0045`
Glauber's salt `rarr Na_(2)SO_(4).10H_(2)O` (M. W. = 322 gm/mol)
Moles of Glauber’s salt dissolved `= (80.575)/(322)"mole" = 0.25 "mole"`
`v_("soln") = 1 dm^(3) = 1l rArr ` Molarity = 0.25 M
Mass of solution `= 1000 xx 1.0772 gm//c c = 1077.2 gm`
Mass of solvent in solution = 1077.2 – 80.575 = 996.625 gm
`:.` 996.625 gm solvent has 0.25 mole solute
Hence, 1000 gm solvent has `= 0.25 xx (1000)/(996.625) = 0.251` molal
`chi_(Na_(2)SO_(4)) = (n_(Na_(2)SO_(4)))/(n_(Na_(2)SO_(4)) + n_(H_(2)O)) = (0.25)/(0.25 + (996.625)/(18))= 0.0045`