When `0.575 xx 10^(-2) kg` of Glaube's salt is dissolved in water, we get `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 the solution.

Answer

A, B, C, D

Explanation

Glauber's salt is `Na_(2)SO_(4).10H_(2)O`.
Given mass `= 8.0575 xx 10^(-2) kg = 80.575 kg`
`Mw = 322 g mol^(-1)`
`Mw` of `Na_(2) SO_(4)` in `80.575 g` sample `= (142)/(322) xx 80.575`
`= 35.53 g`
Molarity `(M) = (35.53)/(142) xx 1 = 0.2502`
Density `= 1077.2 kg m^(-3) = 1077.2 g L^(-1)`
Desity `= ("Mass")/("Volume")`
Mass of solution `= "density" xx "volume"`
`1077.2 xx 1077.2 g`
Molality `(m) = (35.53)/(142) xx (1)/(1041.67) xx 1000 = 0.2402`
Moles of `Na_(2)SO_(4) = (35.53)/(142) = 0.2502`
Moles of `H_(2) O = (1041.67)/(18) = 57.87`
Moles fraction `(Na_(2)SO_(4)) = ("Moles of" Na_(2)SO_(4))/("Total Moles")`
`= (0.2502)/(0.2502 + 57.87)`
`= 4.304 xx 10^(-3)`