Question

When $0.575\times {10}^{-2}kg$ of Glauber's salt ($N{a}_{2}S{O}_4\cdot 10H_{2}O$) is dissolved in $3$ g water, we get $1$ $d{m}^3$ of a solution of density $1077.2kg{m}^{-3}$. Calculate the molarity, molality and mole fraction of $N{a}_{2}S{O}_4$ in the solution:

• A. $Molarity\left(M\right)=0.2502$, $Molality\left(m\right)=0.2402$, $Molefraction=4.304\times {10}^{-3}$
• B. $Molarity\left(M\right)=0.2402$, $Molality\left(m\right)=0.2502$, $Molefraction=4.304\times {10}^{-3}$
• C. $Molarity\left(M\right)=0.2502$, $Molality\left(m\right)=0.2402$, $Molefraction=4.304\times {10}^{-1}$
• D. None of these

Answer 1

The correct option is A $Molarity\left(M\right)=0.2502$, $Molality\left(m\right)=0.2402$, $Molefraction=4.304\times {10}^{-3}$

The formula for Glauber's salt is $N{a}_{2}S{O}_4\cdot 10H_{2}O$.
Given mass $=8.0575\times {10}^{-2}kg=80.575kg$
Molecular weight $=322gmo{l}^{-1}$
Molecular weight of $N{a}_{2}S{O}_4=142gmo{l}^{-1}$
Mass of $N{a}_{2}S{O}_4$ in $80.575$ g sampe $=\frac{142}{322}\times 80.575$ $=35.53g$
Molarity (M) $=\frac{35.52}{142}\times 1=0.2502M$
$Density=1077.2kg{m}^{-3}=1077.2g{L}^{-1}$
$Density=\frac{Mass}{Volume}$
Mass of solution $=density\times volume$ $=1077.2\times 1=1077.2g$
Mass of water $=1077.2-35.53=1041.67g$
$Molality\left(m\right)=\frac{35.53}{142}\times \frac{1}{1041.67}\times 100=0.2402$
Moles of $N{a}_{2}S{O}_4=\frac{35.53}{142}=0.2502$
Moles of $H_{2}O=\frac{1041.67}{18}=57.87$
Moles fraction $\left(N{a}_{2}S{O}_4\right)=\frac{\text{Moles of}N{a}_{2}S{O}_4}{\text{Total Moles}}$ $=\frac{0.2502}{0.2502+57.87}$ $=4.304\times {10}^{-3}$