Browse Questions

# $8.0575 × 10^{–2}$ 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 molarity 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} =$$\large\frac{1077.22\times 1000gm}{10^6cm^3} Volume (V) 1L=1000ml Molarity of Na_2SO_4 ⋅10H_2O =\large\frac{w}{M}\times \frac{1000}{V(in\;ml} \Rightarrow \large\frac{80.575}{322}$$\times \frac{1000}{1000}$
$\Rightarrow 0.25ml\;lit^{-1}$
$\Rightarrow 0.25ml\;dm^{-3}$
$\Rightarrow 0.25M$
Hence (A) is the correct answer.