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.

$(1983,5 M)$

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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:

  1. (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 $Na^{+}=\frac{3 \times 2}{776} \times 100=7.73 m$



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