InterviewSolution
Saved Bookmarks
| 1. |
`0.7 g " of " Na_(2)CO_(3).xH_(2)O` were dissolved in water and the volume was made to `100 mL, 20 mL` of this solution required `19.8 mL " of " N//10 HCl` for complete neutralization. The value of `x` is:A. 7B. 3C. 2D. 5 |
|
Answer» Correct Answer - C Number of milliequivalents of `Na_(2)CO_(3).xH_(2)O` in 20 mL `=19.8xx(1)/(10)` =1.98 `therefore` Number of milliequivalents in 100 mL `=1.98xx5=9.9` `("Mass")/("Equivalent mass")xx1000=9.9` `(0.7)/(M//2)xx1000=9.9` M=141.40 (Molar mass of `Na_(2)CO_(3). xH_(2)O)` 106+18x=141.40 x=2 |
|