Mole Concept and Stoichiometry — Question 35

Mass of 1 mole of NaOH = Na + O + H = 23 + 16 + 1 = 40 g

40 g of NaOH contains 6.022 × 10²³ molecules

∴ 1000 g of NaOH contains = $\dfrac{6.022 \times 10^{23} \times 1000}{40}$

= 1.5 × 10²⁵ molecules