Some Basic Concepts of Chemistry - Result Question 40
47. $8.0575 \times 10^{-2} \mathrm{~kg}$ of Glauber’s salt is dissolved in water to obtain $1 \mathrm{dm}^3$ of solution of density $1077.2 \mathrm{~kg} \mathrm{~m}^{-3}$. Calculate the molality, molarity and mole fraction of $\mathrm{Na}_2 \mathrm{SO}_4$ in solution.
(1994, 3M)
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Answer:
Correct Answer: 47. $\left(4.3 \times 10^{-3}\right)$
Solution:
- Molar mass of Glauber’s salt $\left(\mathrm{Na}_2 \mathrm{SO}_4 \cdot 10 \mathrm{H}_2 \mathrm{O}\right)$
$=23 \times 2+32+64+10 \times 18=322 \mathrm{~g}$
$\begin{aligned} & \Rightarrow \text { Mole of } \mathrm{Na}_2 \mathrm{SO}_4 \cdot 10 \mathrm{H}_2 \mathrm{O} \text { in 1.0 } \mathrm{L} \text { solution }=\frac{80.575}{322}=0.25 \\ & \Rightarrow \text { Molarity of solution }=0.25 \mathrm{M} \\ & \text { Also, weight of } 1.0 \mathrm{~L} \text { solution }=1077.2 \mathrm{~g} \\ & \text { weight of } \mathrm{Na}_2 \mathrm{SO}_4 \text { in } 1.0 \mathrm{~L} \text { solution }=0.25 \times 142=35.5 \mathrm{~g} \\ & \Rightarrow \text { Weight of water in } 1.0 \mathrm{~L} \text { solution }=1077.2-35.5=1041.7 \mathrm{~g} \\ & \Rightarrow \text { Molality }=\frac{0.25}{1041.7} \times 1000=0.24 \mathrm{~m} \\ & \text { Mole fraction of } \mathrm{Na}_2 \mathrm{SO}_4=\frac{\mathrm{Mole} \text { of } \mathrm{Na}_2 \mathrm{SO}_4}{\text { Mole of } \mathrm{Na}_2 \mathrm{SO}_4+\text { Mole of water }} \\ & =\frac{0.25}{0.25+\frac{1041.7}{18}} \\ & =4.3 \times 10^{-3} \text {. } \\ & \end{aligned}$
BETA
