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| 1. |
Amount of NaOHpresent in 200 mlof 0.5Nsolution is |
| Answer» 1N NaOH = 40 G NaOH in 1000 mL of the solution = 1 MOLE of NaOH in 1000 mL of the solution. 0.5 N NaOH = 20 g NaOH in 1000 mL of the solution = 0.5 mole of NaOH in 1000 mL of the solution.pls friend MARK me as brainliest | |