Saved Bookmarks
| 1. |
How much of NaOH is required to neutralise 1500 cm^(3) of 0.1N HCl? (Na= 23) |
|
Answer» 4g E.W (NaOH) = 40M V= 1500 `cm^(3)`, N= 0.1 N, W= ? `rArr W= (1500 xx 0.1 xx 40)/(1000) rArr W= 6gm` |
|