Some Basic Concepts of Chemistry - Result Question 56
####55. The density of a $3 M$ sodium thiosulphate solution $\left(Na _2 S _2 O _3\right)$ is $1.25 g$ per $mL$. Calculate (i) the percentage by weight of sodium thiosulphate (ii) the mole fraction of sodium thiosulphate and (iii) the molalities of $Na^{+}$and $S _2 O _3^{2-}$ ions.
$(1983,5 M)$
Show Answer
Answer:
Correct Answer: 55. (i) 37.92 , (ii) 0.065 , (iii) $7.73 m$ 56. (a) 0.6 , (b) 24
$\begin{array}{lll}\text { 58. (i) } 0.0179 g \text {, (ii) } 10.6 % & \text { 59. }(0.437) & \text { 61. } 20 %\end{array}$
Topic 2
| 1. $(d)$ | 2. $(b)$ | 3. (a) | 4. (a) |
|---|---|---|---|
| 5. $(d)$ | 6. ( $^{*}$ | 7. (b) | 8. (c) |
| 9. (d) | 10. $(c)$ | 11. (a) | 12. (d) |
| 13. $(b)$ | 14. $(c)$ | 15. (a) | 16. (d) |
| 17. (a) | 18. $(b)$ | 19. (a) | 20. (a) |
| 21. $(b)$ | 22. $(c)$ | 23. (b) | 24. (a, b, d) |
| 25. $(2992)$ | 26. (b) | 27. $7 / 3$ | 28. $(5)$ |
| 29. $(2)$ | 30. $(3)$ | 31. $(1008 g)$ | 33. $(8.096 mL)$ |
| 34. $(0.062 M)$ | 35. $(1.334 V)$ | 39. $(85 %)$ | 41. $\left(1.04 \times 10^{4}\right)$ |
| 42. $(1: 2)$ | 45. $(16.67 mL)$ | 46. $\left(6.5 gL^{-1}\right)$ | 47. $(6.5376 g)$ |
| 48. $(2)$ | 49. $(Ca)$ |
Solution:
- (a) Let us consider $1.0 L$ solution for all the calculation.
(i) Weight of $1 L$ solution $=1250 g$
Weight of $Na _2 S _2 O _3=3 \times 158=474 g$
$\Rightarrow$ Weight percentage of $Na _2 S _2 O _3=\frac{474}{1250} \times 100=37.92$
(ii) Weight of $H _2 O$ in $1 L$ solution $=1250-474=776 g$
Mole fraction of $Na _2 S _2 O _3=\frac{3}{3+\frac{776}{18}}=0.065$
(iii) Molality of $Na^{+}=\frac{3 \times 2}{776} \times 100=7.73 m$
