Question #76674
how many moles of NaOH are present in 18.5 ml of 0.130 M NaOH
Expert's answer
СM = n/V n = CM · V
n (NaOH) = 0.130 · 18.5 = 2.405 mmol = 0.002 mol
n (NaOH) = 0.130 · 18.5 = 2.405 mmol = 0.002 mol
