Question

45 grams of ammonia, NH3, in 0.75 L of solution.

Answer 1

2.3 moles of sodium chloride in 0.45 liters of solution . necessarile correct ... 0.025 L. (1.3 SF). 45 grams of ammonia in 0.75 L of solution. 45g x 1 mol = 2. demol ... anol x 1,50 = 3 mol NaOH x 40,019 = 120g Naolt.