# $8.0575\times 10^{-2}kg$ of Glauber's salt is dissolved in water.Obtain $1clm^3$ of a solution of density $1077.2kgm^{-3}$.Calculate the molarity and the mole fraction of $Na_2SO_4$ in solution .

$(a)\;0.2502m,0.24m,4.3\times 10^{-3}\\(b)\;0.3m,0.1m,2\times 10^{-2}\\(c)\;.15m,0.3m,1.3\times 10^{-3}\\(d)\;.3m,0.156m,1.3\times 10^{-2}$

Step 1:
Mole weight $Na_2SO_4.10H_2O=322$
Weight of $Na_2SO_4$ in $8.0575\times 10^{-2}kg$ glauber salt
$\Rightarrow \large\frac{142\times 8.0575\times 10^{-2}}{322}$
$\Rightarrow 3.5533\times 10^{-2}kg$
M of $Na_2SO_4=\large\frac{3.5533\times 10^{-2}}{142\times 10^{-3}\times 1}$
$\Rightarrow 0.2502m$
Step 2:
Molality of $Na_2SO_4=\large\frac{3.5533\times 10^{-2}}{142\times 10^{-3}\times 1}$
$\Rightarrow \large\frac{3.5533\times 10^{-2}}{142\times 10^{-3}\times \Large\frac{1041.667}{10^{-3}}}$
$\Rightarrow 0.24m$
Step 3:
Mole fraction of $Na_2SO_4=\large\frac{Mole\;of\;Na_2SO_4}{Mole\;of\;Na_2SO_4+Mole\;of\;H_2O}$
$\Rightarrow \large\frac{3.5533\times 10^{-2}/142\times 10^{-3}}{3.5533\times 10^{-2}/142\times 10^{-3}+1041.667/18}$
$\Rightarrow 4.3\times 10^{-3}$
Hence (a) is the correct answer.