`40.25g` of Glauber's salt `(Na_(2)SO_(4).10H_(2)O)` is dissolved in water to obtain `500mL` of solution of density `1077.5gdm^(-3)`. The molality of `Na_(2)SO_(4)` in solution is about:
A
`0.48molkg^(-1)`
B
`0.24molkg^(-1)`
C
`0.12molkg^(-1)`
D
`0.84molkg^(-1)`
Text Solution
Verified by Experts
The correct Answer is:
B
Molar mass of Glauber's salt `Na_(2)SO_(4)10H_(2)O=322mol^(-1)` Molar mass of `Na_(2)SO_(4)=142gmol^(-1)` Mass of `Na_(2)SO_(4)` in solution `=(40.25 g) ((142"g mol"^(-1))/(322 "gmol"^(-1)))=17.75 g` Mass of solution `=(500mL)(1.0775gmol^(-1))=538.75g` Mass of water`=(538.75--17.75)g=521g` Amount of `Na_(2)SO_(4)=(17.55g)/(142gmol^(-1))=0.125mol` Molality of `Na_(2)SO_(4)=(0.125mol)/(0.521kg)=0.24molkg^(-1)`
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