`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) 0.24 m (iii) 0.0043

Explanation

Mol .mass of Glauber .s salt `(Na_(2)SO_(4) .10H_(2)O) = 322`
Mol.mass of `Na_(2)SO_(4) = 142`
Wt. of `Na_(2)SO_(4) = (142 xx 8.0575 xx 10^(-2) xx 10^(3))/(322)`
` = 35.533g`
Moles of `Na_(2)SO_(4) = (35.533)/(142) = 0.25 M`
(i) Molarity ` = (0.25)/(1) = 0.25 M`
(ii) Density` = (1077.2 Kg)/(m^(3)) =(1077 .2 xx 10^(3) g)/( 10^(6)cm^(3)) = 1.0722 g cm^(-3)`
Mass of `1 dm^(3)` of solution ` = 1000 xx 1.0772 = 1077.2 g`
Mass of water ` = 1077.2 - 35.3 = 1041.7 g`
Molality ` = (0.25)/(1041.7) xx 100 = 0.24 m`
(iii) Moles of water ` =(1041.7)/(18) = 51.87`
Moles fraction of `Na_(2)SO_(4) = (0.25)/(0.25 + 57.87) = 4.3 xx 10^(-3)`.