SATHEE: Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry - Result Question 56

####55. The density of a $3 M$ sodium thiosulphate solution $(N a_{2} S_{2} O_{3})$ is $1.25 g$ per $m L$ . Calculate (i) the percentage by weight of sodium thiosulphate (ii) the mole fraction of sodium thiosulphate and (iii) the molalities of $N a^{+}$ 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

$Missing \end{array}$

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. $(8.096 m L)$
34. $(0.062 M)$	35. $(1.334 V)$	39. $(85$	41. $(1.04 \times 10^{4})$
42. $(1 : 2)$	45. $(16.67 m L)$	46. $(6.5 g L^{- 1})$	47. $(6.5376 g)$
48. $(2)$	49. $(C a)$

Solution:

(a) Let us consider $1.0 L$ solution for all the calculation.

(i) Weight of $1 L$ solution $= 1250 g$

Weight of $N a_{2} S_{2} O_{3} = 3 \times 158 = 474 g$

$\Rightarrow$ Weight percentage of $N a_{2} S_{2} O_{3} = \frac{474}{1250} \times 100 = 37.92$

(ii) Weight of $H_{2} O$ in $1 L$ solution $= 1250 - 474 = 776 g$

Mole fraction of $N a_{2} S_{2} O_{3} = \frac{3}{3 + \frac{776}{18}} = 0.065$

(iii) Molality of $N a^{+} = \frac{3 \times 2}{776} \times 100 = 7.73 m$

Some Basic Concepts of Chemistry - Result Question 56

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Some Basic Concepts of Chemistry - Result Question 56

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu