Glauber's salt is $\mathrm{Na}_2 \mathrm{SO}_4 \cdot 10 \mathrm{H}_2 \mathrm{O}$.

Molecular weight of $\mathrm{Na}_2 \mathrm{SO}_4=142$

Molecular weight of $\mathrm{Na}_2 \mathrm{SO}_4 \cdot 10 \mathrm{H}_2 \mathrm{O}=322$

Weight of Glauber's salt taken $=8.0575 \times 10^{-2} \mathrm{~kg}=80.575 \mathrm{~g}$

Weight of anhydrous $\mathrm{Na}_2 \mathrm{SO}_4$ in $322 \mathrm{~g}$ of Glauber's salt

$$=142 / 322 \mathrm{~g}$$

$\therefore$ Weight of anhydrous $\mathrm{Na}_2 \mathrm{SO}_4$ in $80.575 \mathrm{~g}$ of Glauber's salt

$$ =\frac{142}{322} \times 80.575=35.53 \mathrm{~g} $$

Hence number of moles of $\mathrm{Na}_2 \mathrm{SO}_4$ per $\mathrm{dm}^3$ of solution

$$ =\frac{35.53}{142}=0.25 $$

So, the molarity of solution is $0.25 \mathrm{M}$

$$ \begin{aligned} \text { Density of solution } & =1077.2 \mathrm{~kg} \mathrm{~m}^{-3} \\\\ & =\frac{1077.2 \times 10^3}{10^6} \mathrm{~g} \mathrm{~cm}^{-3}=1.0772 \mathrm{~g} \mathrm{~cm}^{-3} \end{aligned} $$

$\begin{aligned} & \text { Weight of solution }=\text { volume } \times \text { density } \\\\ & =1000 \times 1.0772 \mathrm{~g}=1077.2 \mathrm{~g} \\\\ & \text { Weight of water }=(1077.2-35.53)=1041.67 \mathrm{~g} \\\\ & \text { Molality of solution }=\frac{0.25}{1041.67} \times 1000=0.24 \mathrm{~m}\end{aligned}$

$\begin{aligned} & \text { Number of moles of water in solution }=\frac{1041.67}{18}=57.87 \\\\ & \text { Mole fraction of } \mathrm{Na}_2 \mathrm{SO}_4=\frac{\text { Moles of } \mathrm{Na}_2 \mathrm{SO}_4}{\text { Total number of moles }} \\\\ & \qquad=\frac{0.25}{0.25+57.87}=0.0043 \text { or } 4.3 \times 10^{-3}\end{aligned}$