Solutions and Colligative Properties - Result Question 69

####40. $M X _2$ dissociates into $M^{2+}$ and $X^{-}$ions in an aqueous solution, with a degree of dissociation $(\alpha)$ of 0.5 . The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is

(2014 Adv.)

1. (c) 2. (c) 3. (c) 4. (b)
5. (d) 6. (c) 7. (c) 8. (b)
9. (c) 10. (d) 11. (c) 12. (d)
13. (d) 14. (b) 15. (a) 16. (a)
17. (a) 18. (a) 19. (c) 20. (a)
21. (a) 22. (b) 23. (d) 24. (a)
25. (a) 26. (a, d) 27. $(0.05)$ 28. $(75 %)$
30. $\left(0.23^{\circ} C\right)$ 33. $(23.44 mm)$ 34. $(156 g / mol)$ 35. (d)
36. (a) 37. (b) 38. $\left(K _f\right)$ 40. (2)
Show Answer

Solution:

  1. $M X _2 \longrightarrow M^{2+}+2 X^{-}$

van’t Hoff factor for any salt can be calculated by using equation $i=1+\alpha(n-1)$

where, $n=$ number of constituent ions

$$ \begin{aligned} & \therefore \quad i\left(M X _2\right)=1+\alpha(3-1)=1+2 \alpha \ & \frac{\left(\Delta T _f\right) _{\text {observed }}}{\left(\Delta T _f\right) _{\text {theoretical }}}=i=1+2 \alpha \ & \therefore \quad i=1+2 \times 0.5 \Rightarrow i=2 \end{aligned} $$

Download Chapter Test http://tinyurl.com/yxr5edmo



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक