`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

`(10.42)`

Explanation

Molar mass of Glauber's salt `(Na_(2)SO_(4).10H_(2)O)`
`= 23 xx 2+32 + 64 + 10 xx 18 = 322 g`
`rArr` Mole of `Na_(2)SO_(4).10H_(2)O` in `1.0 L` solution `= (80.575)/(322) = 0.25`
`rArr` Molarity of solution `= 0.25 M`
Also, weight of `1.0 L` solution `= 1077.2 g`
weight of `Na_(2)SO_(4)` in `1.0 L` solution `= 0.25 xx 142 = 35.5 g`
`rArr` Weight of water in `1.0 L` solution `= 1077.2 -35.5 =1041.7 g`
`rArr` Molarity `= (0.25)/(1041.7) xx 1000 = 0.24 m`
Mole fraction of `Na_(2)SO_(4) = ("Mole of" Na_(2)SO_(4))/("Mole of" Na_(2)SO_(4)+"Mole of water")`
`= (0.25)/(0.25 + (1041.7)/(18))`
`= 4.3 xx 10^(-3)`.