`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.

Explanation

`W_(1)`= Wt. of salt `=8.0575xx10^(-2)kg`
`=80.575g`
d= density of the solution `1077.2 kgm^(-3)=1.0772 g cm^(3)`
Volume `(V)= 1 dm^(3)=1l|1000m|`
Molarity (M) of `Na_(2)SO_(4).10H_(2)O=(1000xxw_(1))/(m_(1)xxv)`
`=(100xx80.575)/(322xx1000)=0.25 "moldm"^(-3)`
Molaliry of `Na_(2)SO_(4)(m)=(1000xxM)/((1000xxd-Mxxm_(1)))`
`=(1000xx0.25)/((1000xx1.0772-0.25xx142))`
`=0.24 "molkg"^(-1)` solvent.
Here `m_(1)=142 "gmol"^(-1)` for anyhdrous salt `Na_(2)SO_(4)`
AS `m=(1000xxx_(1))/((1-x_(1))m_(2))`
`:.(x_(1))/((1-x_(1)))=(mxxm_(2))/(1000)=(0.24xx18)/(1000)`
Mole of fraction of `Na_(2)SO_(4)=0.0043`