Hi Reese, ----------- 1 mole gram of NaOH weight 40 g"r (Na = 23, O = 16, H = 1...... 23 + 16 + 1 = 40). So in 20.1 g"r there are 20.1 : 40 = 0.5025 mole-g"r. Each mole-g"r (of any compound) contains N.A. (Avogadro number [ = 6.023 * 10^23]) moles..... So in 0.5025 mole-g"r there are 0.5025 * 6 ...