# $8.0575 × 10^{–2}$ kg of Glauber’s salt ($Na_2SO_4.10H_2O$) is dissolved in water to obtain 1 $dm^3$ of a solution of density $1077.2 kgm^{–3}$. Calculate the molality fraction of $Na_2SO_4$ in the solution.

Wt. of Glauber’s salt $(w_1) = 8.057 \times 10^{–2} kg = 80.575 g$
Density of solution (d) = $1077.2 kg m^{–3} =$$\large\frac{1077.22\times 1000gm}{10^6cm^3} Volume (V) 1L=1000ml Molality of Na_2SO_4 =$$\large\frac{1000M}{1000d-MM_1}$
$\Rightarrow \large\frac{1000\times 0.25}{1000\times 1.0722-0.25\times 142}$
$\Rightarrow 0.24molkg^{-1}\approx 0.25m$
Hence (A) is the correct answer.