Some Basic Concepts of Chemistry - Result Question 47
####47. $8.0575 \times 10^{-2} kg$ of Glauber’s salt is dissolved in water to obtain $1 dm^{3}$ of solution of density $1077.2 kg m^{-3}$. Calculate the molality, molarity and mole fraction of $Na _2 SO _4$ in solution.
(1994, 3M)
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Solution:
- Molar mass of Glauber’s salt $\left(Na _2 SO _4 \cdot 10 H _2 O\right)$
$$ =23 \times 2+32+64+10 \times 18=322 g $$
$\Rightarrow$ Mole of $Na _2 SO _4 \cdot 10 H _2 O$ in $1.0 L$ solution $=\frac{80.575}{322}=0.25$
$\Rightarrow$ Molarity of solution $=0.25 M$
Also, weight of $1.0 L$ solution $=1077.2 g$
weight of $Na _2 SO _4$ in $1.0 L$ solution $=0.25 \times 142=35.5 g$
$\Rightarrow$ Weight of water in $1.0 L$ solution $=1077.2-35.5=1041.7 g$
$\Rightarrow$ Molality $=\frac{0.25}{1041.7} \times 1000=0.24 m$
Mole fraction of $Na _2 SO _4=\frac{\text { Mole of } Na _2 SO _4}{\text { Mole of } Na _2 SO _4+\text { Mole of water }}$
$$ \begin{aligned} & =\frac{0.25}{0.25+\frac{1041.7}{18}} \ & =4.3 \times 10^{-3} . \end{aligned} $$
