250 mL of 0 . 5 M NaOH was added to 500 mL of 1 M HCl . The number of unreacted HCl molecules in the solution after the complete reaction is p × 10 21 . Find out p (Nearest integer) N A = 6 . 022 × 10 23

Step-by-step solution

We known that no. of moles = V litre  × Molarity & No. of millimoles = V ml × Molarity so millimoles of NaOH = 250 × 0 . 5 = 125 Millimoles of HCl = 500 × 1 = 500 Now reaction is so millimoles of HCl left = 375 Moles of HCl = 375 × 10 - 3 No. of HCl molecules = 6 . 022 × 10 23 × 375 × 10 - 3 = 225 . 8 × 10 21 ≃ 226 × 10 21 = 226