QJEEM21Ch701f: JEE Main 2021 Chemistry - Class 11 JEEM21Ch701f Solution: 250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number …

We know that, number of moles = V_L $$\times$$ molarity and number of millimoles = V_mL $$\times$$ molarity

So, millimoles of NaOH = 250 $$\times$$ 0.5 = 125

Millimoles of HCl = 500 $$\times$$ 1 = 500

Now, reaction is


125 millimoles of NaOH reacts with 125 millimoles of HCl. So, millimoles of HCl left = 375

Moles of HCl = 375 $$\times$$ 10^$$-$$3

Number of HCl molecules

= Avogadro's constant (N_A) $$\times$$ moles of HCl

= 6.022 $$\times$$ 10²³ $$\times$$ 375 $$\times$$ 10^$$-$$3

= 225.8 $$\times$$ 10²¹ = 226 $$\times$$ 10²¹

Therefore, answer is 226.